cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A262951 a(1) = 1, a(2) = 3, a(3) = 4 and for n>=4, a(n) = (a(n-3)+a(n-2)+a(n-1)+k) mod 10 where k = a(n/6) if n is divisible by 6, else 0.

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%I A262951 #10 Oct 19 2024 15:57:32
%S A262951 1,3,4,8,5,7,0,2,9,1,2,5,8,5,8,1,4,7,2,3,2,7,2,9,8,9,6,3,8,2,3,3,8,4,
%T A262951 5,4,3,2,9,4,5,8,7,0,5,2,7,6,5,8,9,2,9,9,0,8,7,5,0,3,8,1,2,1,4,9,4,7,
%U A262951 0,1,8,4,3,5,2,0,7,7,4,8,9,1,8,3,2,3,8
%N A262951 a(1) = 1, a(2) = 3, a(3) = 4 and for n>=4, a(n) = (a(n-3)+a(n-2)+a(n-1)+k) mod 10 where k = a(n/6) if n is divisible by 6, else 0.
%C A262951 This sequence is similar to A130893. Every term of index k is the sum of the 3 preceding terms modulo 10, except that for every sixth term the sum includes also the term of index k/6.
%C A262951 Lambert gave this sequence in Anlage zur Architectonic as a kind of early pseudorandom sequence. - _Charles R Greathouse IV_, Oct 05 2015
%H A262951 Maarten Bullynck, <a href="http://images-archive.math.cnrs.fr/L-histoire-de-l-informatique-et-l-histoire-des-mathematiques-rencontres.html">L’histoire de l’informatique et l’histoire des mathématiques : rencontres, opportunités et écueils</a>, Images des Mathématiques, CNRS, 2015 (in French).
%H A262951 Johann Heinrich Lambert, <a href="http://www.deutschestextarchiv.de/book/show/lambert_architectonic01_1771">Anlage zur Architectonic, oder Theorie des Einfachen und des Ersten in der philosophischen und mathematischen Erkenntniß</a>, 1771.
%H A262951 <a href="/index/Ps#PRN">Index entries for sequences related to pseudo-random numbers.</a>
%F A262951 a(n) = (a(n-3) + a(n-2) + a(n-1)) mod 10 if n is not a multiple of 6.
%F A262951 a(n) = (a(n-3) + a(n-2) + a(n-1) + a(n/6)) mod 10 if n is a multiple of 6.
%e A262951 a(6) = 4+8+5 = (17 + a(6/6)) mod 10 = (17 + 1) mod 10 = 8.
%o A262951 (PARI) lista(nn) = {va = vector(nn); va[1] = 1; va[2] = 3; va[3] = 4; for (k=4, nn, va[k] = va[k-3] + va[k-2] + va[k-1]; if (! (k % 6) && (k > 6), va[k] += va[k/6]); va[k] = va[k] % 10;); va;}
%Y A262951 Cf. A130893.
%K A262951 nonn
%O A262951 1,2
%A A262951 _Michel Marcus_, Oct 05 2015