Original entry on oeis.org
1, 9, 49, 249, 1249, 6249, 31249, 156249, 781249, 3906249, 19531249, 97656249, 488281249, 2441406249, 12207031249, 61035156249, 305175781249, 1525878906249, 7629394531249, 38146972656249, 190734863281249
Offset: 0
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[2*5^n-1: n in [0..30]]; // Vincenzo Librandi, Aug 10 2013
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CoefficientList[Series[(1 + 3 x) / ((1 - 5 x) (1 - x)), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 10 2013 *)
NestList[5#+4&,1,30] (* Harvey P. Dale, Jul 04 2014 *)
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a(n)=2*5^n-1 \\ Charles R Greathouse IV, Sep 24 2015
A094424
Array read by antidiagonals: Solutions to Schmidt's Problem.
Original entry on oeis.org
1, 1, 1, 1, 2, 1, 1, 4, 10, 1, 1, 8, 68, 56, 1, 1, 16, 424, 1732, 346, 1, 1, 32, 2576, 48896, 51076, 2252, 1, 1, 64, 15520, 1383568, 6672232, 1657904, 15184, 1, 1, 128, 93248, 39776000, 873960976, 1022309408, 57793316, 104960, 1, 1, 256, 559744, 1159151680, 116758856608, 615833930816, 176808084544, 2117525792, 739162, 1
Offset: 1
Array begins:
1, 1, 1, 1, 1, 1, ...
1, 2, 10, 56, 346, 2252, ...
1, 4, 68, 1732, 51076, 1657904, ...
1, 8, 424, 48896, 6672232, 1022309408, ...
1, 16, 2576, 1383568, 873960976, 615833930816, ...
1, 32, 15520, 39776000, 116758856608, 371558588978432, ...
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eq[r_, n_] := eq[r, n] = Sum[Binomial[n, k]^r*Binomial[n + k, k]^r, {k, 0, n}] == Sum[Binomial[n, k]*Binomial[n + k, k]*t[r, k], {k, 0, n}]; c[r_, k_] := t[r, k] /. Solve[Table[eq[r, n], {n, 0, k}], t[r, k]] // First; lg = 10; m = Table[c[r, k], {r, 1, lg}, {k, 0, lg - 1}];
Flatten[ Table[ Reverse @ Diagonal[ Reverse /@ m, k],{k, lg - 1, -lg + 1, -1}]][[1 ;; 55]] (* Jean-François Alcover, Jul 20 2011 *)
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A094424row(r,kmax)={ local(nmat,rhs,cv) ; nmat=matrix(kmax+1,kmax+1) ; rhs=matrix(kmax+1,1) ; for(n=0,kmax, for(k=0,kmax, nmat[n+1,k+1]=binomial(n,k)*binomial(n+k,k) ; ) ; rhs[n+1,1]=sum(i=0,n,binomial(n,i)^r*binomial(n+i,i)^r) ; ) ; cv=matsolve(nmat,rhs) ; } A094424(nmax)={ local(T,c) ; T=matrix(nmax,nmax) ; for(r=1,nmax, c=A094424row(r,nmax-1) ; for(i=1,nmax, T[r,i]=c[i,1] ; ) ; ) ; return(T) ; } { rmax=10 ; T=A094424(rmax) ; for(d=0,rmax-1, for(c=0,d, print1(T[d-c+1,c+1],",") ; ) ; ) ; } \\ R. J. Mathar, Oct 06 2006
Original entry on oeis.org
1, 11, 89, 659, 4721, 33371, 234569, 1644899, 11523041, 80687531, 564891449, 3954476339, 27682042961, 193776426491, 1356441362729, 9495108670979, 66465818092481, 465260898834251, 3256826808400409, 22797789208484819
Offset: 0
A385178
Triangle T(n,k) read by rows in which the n-th diagonal lists the n-th differences of A001047, 0 <= k <= n.
Original entry on oeis.org
0, 1, 1, 3, 4, 5, 7, 10, 14, 19, 15, 22, 32, 46, 65, 31, 46, 68, 100, 146, 211, 63, 94, 140, 208, 308, 454, 665, 127, 190, 284, 424, 632, 940, 1394, 2059, 255, 382, 572, 856, 1280, 1912, 2852, 4246, 6305, 511, 766, 1148, 1720, 2576, 3856, 5768, 8620, 12866, 19171
Offset: 0
Triangle begins:
0;
1, 1;
3, 4, 5;
7, 10, 14, 19;
15, 22, 32, 46, 65;
31, 46, 68, 100, 146, 211;
63, 94, 140, 208, 308, 454, 665;
127, 190, 284, 424, 632, 940, 1394, 2059;
255, 382, 572, 856, 1280, 1912, 2852, 4246, 6305;
511, 766, 1148, 1720, 2576, 3856, 5768, 8620, 12866, 19171;
...
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/* As triangle */ [[2^(n-k)*3^k - 2^k : k in [0..n]]: n in [0..9]]; // Vincenzo Librandi, Jun 27 2025
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T:= proc(n,k) option remember;
`if`(n=k, 3^n-2^n, T(n, k+1)-T(n-1, k))
end:
seq(seq(T(n, k), k=0..n), n=0..10); # Alois P. Heinz, Jun 24 2025
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t[n_, 0] := 3^n - 2^n; t[n_, k_] := t[n, k] = t[n + 1, k - 1] - t[n, k - 1]; Table[t[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, Jun 20 2025 *)
Showing 1-4 of 4 results.
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