Rational generating functions for the Certain Stanley-Stern Sums By Shalosh B. Ekhad Theorem Number, 1 --------------------------------- Let Z[n] be the integer sequence whose generating function is infinity ----- 2 \ j -t + 1 ) Z[j] t = ---------------- / 3 2 ----- -t - t - t + 1 j = 0 Let n - 1 --------' ' | | F[n](x) = | | | | | | i = 0 Z[i + 2] (Z[i + 1] + Z[i + 2]) (Z[i] + Z[i + 1] + Z[i + 2]) (1 + x + x + x ) Write: infinity ----- \ i F[n](x) = ) a(n, i) x / ----- i = 0 Let : infinity ----- \ H(n) = ) a(n, k) / ----- k = 0 Then infinity ----- \ n ) H(n) t / ----- n = 0 equals (I-Mt)^(-1) v [1] where M is a certain square matrix of dimension, 1 and v is a certain vector of length, 1 that are too big to display. At any rate we can use them to find the first, 31, terms starting at n=0, for the sake of the OEIS. [1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1048576, 4194304, 16777216, 67108864, 268435456, 1073741824, 4294967296, 17179869184, 68719476736, 274877906944, 1099511627776, 4398046511104, 17592186044416, 70368744177664, 281474976710656, 1125899906842624, 4503599627370496, 18014398509481984, 72057594037927936, 288230376151711744, 1152921504606846976] ----------------------------- This took, 0.012, seconds. Theorem Number, 2 --------------------------------- Let Z[n] be the integer sequence whose generating function is infinity ----- 2 \ j -t + 1 ) Z[j] t = ---------------- / 3 2 ----- -t - t - t + 1 j = 0 Let n - 1 --------' ' | | F[n](x) = | | | | | | i = 0 Z[i + 2] (Z[i + 1] + Z[i + 2]) (Z[i] + Z[i + 1] + Z[i + 2]) (1 + x + x + x ) Write: infinity ----- \ i F[n](x) = ) a(n, i) x / ----- i = 0 Let : infinity ----- \ 2 H(n) = ) a(n, k) / ----- k = 0 Then infinity ----- \ n ) H(n) t / ----- n = 0 equals (I-Mt)^(-1) v [1] where M is a certain square matrix of dimension, 60 and v is a certain vector of length, 60 that are too big to display. At any rate we can use them to find the first, 31, terms starting at n=0, for the sake of the OEIS. [1, 4, 34, 288, 2456, 21460, 186734, 1625388, 14166464, 123430006, 1075474750, 9371493884, 81659762250, 711558410054, 6200320596352, 54027786655086, 470782929541588, 4102270099387924, 35746024448781384, 311480828531905204, 2714156557683099148, 23650398385012463124, 206082937999357567880, 1795748924967875906740, 15647652452547799778866, 136349254952617608696354, 1188109168716883485416040, 10352850087065923541591616, 90211832191635305237601052, 786080605930016057014231446, 6849685955227362525231262596] ----------------------------- This took, 0.314, seconds. Theorem Number, 3 --------------------------------- Let Z[n] be the integer sequence whose generating function is infinity ----- 2 \ j -t + 1 ) Z[j] t = ---------------- / 3 2 ----- -t - t - t + 1 j = 0 Let n - 1 --------' ' | | F[n](x) = | | | | | | i = 0 Z[i + 2] (Z[i + 1] + Z[i + 2]) (Z[i] + Z[i + 1] + Z[i + 2]) (1 + x + x + x ) Write: infinity ----- \ i F[n](x) = ) a(n, i) x / ----- i = 0 Let : infinity ----- \ 3 H(n) = ) a(n, k) / ----- k = 0 Then infinity ----- \ n ) H(n) t / ----- n = 0 equals (I-Mt)^(-1) v [1] where M is a certain square matrix of dimension, 888 and v is a certain vector of length, 888 that are too big to display. At any rate we can use them to find the first, 31, terms starting at n=0, for the sake of the OEIS. [1, 4, 82, 1444, 26362, 506674, 9582310, 181753786, 3457995082, 65709535264, 1248986949298, 23743974361384, 451349042694418, 8579960440270126, 163102588995472240, 3100516150447008586, 58939806005631843832, 1120426868144274003100, 21298947612102955192540, 404886183956063847689278, 7696757149821012545527216, 146312891025049367377800928, 2781361379288213843498490166, 52872792987098895967855052704, 1005094930985638474061148190450, 19106534152415971112793668096698, 363209122763832547256615103290440, 6904489622748078411066795912629008, 131252146430429268939439553395386376, 2495061458776779094894642565231488894, 47430322872303004814848469272828462675] ----------------------------- This took, 63.620, seconds. Theorem Number, 4 --------------------------------- Let Z[n] be the integer sequence whose generating function is infinity ----- 2 \ j -t + 1 ) Z[j] t = ---------------- / 3 2 ----- -t - t - t + 1 j = 0 Let n - 1 --------' ' | | F[n](x) = | | | | | | i = 0 Z[i + 2] (Z[i + 1] + Z[i + 2]) (Z[i] + Z[i + 1] + Z[i + 2]) (1 + x + x + x ) Write: infinity ----- \ i F[n](x) = ) a(n, i) x / ----- i = 0 Let : infinity ----- \ 4 H(n) = ) a(n, k) / ----- k = 0 Then infinity ----- \ n ) H(n) t / ----- n = 0 equals (I-Mt)^(-1) v [1] where M is a certain square matrix of dimension, 7245 and v is a certain vector of length, 7245 that are too big to display. At any rate we can use them to find the first, 31, terms starting at n=0, for the sake of the OEIS. [1, 4, 214, 7656, 300596, 12781456, 524737850, 21715566888, 903505652300, 37480334152342, 1556159895602026, 64626570162757532, 2683324989411362742, 111422585414399682362, 4626776274862989240736, 192121736818114473034314, 7977706382907613108711900, 331268428648506049923213076, 13755661608167583545562684096, 571193442013263074895897899584, 23718379482624827881035208018156, 984887836261445629308226493731908, 40896727021872872122852017154838300, 1698205892713302329875046055973114852, 70516724781842116967217030979962309046, 2928154058028011160905724931602608935422, 121589399500523843640880609132538691724536, 5048908557454461257565251736085262053342920, 209652138387185284969077336979363205157479092, 8705647695732900230749927299492436550184890634, 361495486780554782930857464378779437028705752636] ----------------------------- This took, 6055.795, seconds. ----------------------------------------- This concludes this article that took, 10581.247, seconds to produce.