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.
%I A178572 #33 Sep 08 2022 08:45:54
%S A178572 11,47,108,194,305,441,602,788,999,1235,1496,1782,2093,2429,2790,3176,
%T A178572 3587,4023,4484,4970,5481,6017,6578,7164,7775,8411,9072,9758,10469,
%U A178572 11205,11966,12752,13563,14399,15260,16146,17057,17993,18954,19940,20951
%N A178572 Numbers with ordered partitions that have periods of length 5.
%C A178572 From each ordered partition of the numbers (10+j) with 0<j<5 one removes the first part z(1) and adds 1 to the next z(1) parts to get a new partition until a period is reached.
%C A178572 The a(n) sequence begins with 11 and each member has 1 period; the b(n) = A022282(n) sequence begins with 12 and each member has 2 periods; the c(n) = A022283(n) sequence begins with 13 and each member has 2 periods; the d(n) = n*(25*n + 3)/2 sequence begins with 14 and each member has 1 period of length 5.
%H A178572 G. C. Greubel, <a href="/A178572/b178572.txt">Table of n, a(n) for n = 1..1000</a>
%H A178572 Amelia Carolina Sparavigna, <a href="https://doi.org/10.5281/zenodo.3470205">The groupoid of the Triangular Numbers and the generation of related integer sequences</a>, Politecnico di Torino, Italy (2019).
%H A178572 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A178572 G.f. for a(n): (11 + 14*x)/(1-x)^3.
%F A178572 for b(n): (12 + 13*x)/(1-x)^3.
%F A178572 for c(n): (13 + 12*x)/(1-x)^3.
%F A178572 for d(n): (14 + 11*x)/(1-x)^3.
%F A178572 All sequences have the same recurrence
%F A178572 s(n+3) = 3*s(n+2) - 3*s(n+1) + s(n)
%F A178572 with s(0)=0, s(1) = 10 + j, s(2) = 45 + 2*j and 0<j<5.
%F A178572 s(n) = n*(25*n - 5 + 2*j)/2 and 0<j<5.
%F A178572 The general formula for numbers with periods of length k: a(k,j,n) = n*(k^2*n - k + 2*j)/2 with 0<j<k.
%F A178572 For j=1 and j=(k-1) the numbers have 1 period.
%F A178572 For 1<j<(k-1) the numbers have A092964(k-4,j-1) periods.
%F A178572 G.f.: (binomial(k,2)*(1+x) + j + (k-j)*x)/(1-x)^3.
%e A178572 For n=11 the period is [(4,3,2,1,1), (4,3,2,2), (4,3,3,1), (4,4,2,1), (5,3,2,1)].
%e A178572 For n=47 the period is [(9,8,7,6,6,4,3,2,1,1), (9,8,7,7,5,4,3,2,2), (9,8,8,6,5,4,3,3,1), (9,9,7,6,5,4,4,2,1), (10,8,7,6,5,5,3,2,1)].
%e A178572 For n=12 the 2 periods are [(4,3,2,2,1), (4,3,3,2), (4,4,3,1), (5,4,2,1), (5,3,2,1,1)] and [(4,3,3,1,1), (4,4,2,2), (5,3,3,1), (4,4,2,1,1), (5,3,2,2)].
%e A178572 For n=49 the 2 periods are [(9,8,7,7,6,4,3,2,2,1), (9,8,8,7,5,4,3,3,2), (9,9,8,6,5,4,4,3,1), (10,9,7,6,5,5,4,2,1), (10,8,7,6,6,5,3,2,1,1)] and [(9,8,8,6,6,4,3,3,1,1), (9,9,7,7,5,4,4,2,2),(10,8,8,6,5,5,3,3,1), (9,9,7,6,6,4,4,2,1,1), (10,8,7,7,5,5,3,2,2)].
%t A178572 LinearRecurrence[{3,-3,1},{11,47,108},50] (* _Harvey P. Dale_, Jan 14 2019 *)
%t A178572 Table[n*(25*n-3)/2, {n,1,50}] (* _G. C. Greubel_, Jan 30 2019 *)
%o A178572 (PARI) a(n)=n*(25*n-3)/2 \\ _Charles R Greathouse IV_, Jun 18 2017
%o A178572 (Magma) [n*(25*n-3)/2: n in [1..50]]; // _G. C. Greubel_, Jan 30 2019
%o A178572 (Sage) [n*(25*n-3)/2 for n in (1..50)] # _G. C. Greubel_, Jan 30 2019
%o A178572 (GAP) List([1..50], n -> n*(25*n-3)/2); # _G. C. Greubel_, Jan 30 2019
%Y A178572 Cf. A092964, A037306.
%K A178572 nonn,easy
%O A178572 1,1
%A A178572 _Paul Weisenhorn_, Dec 24 2010