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 A104698 #50 May 27 2025 03:29:20
%S A104698 1,2,1,3,4,1,4,9,6,1,5,16,19,8,1,6,25,44,33,10,1,7,36,85,96,51,12,1,8,
%T A104698 49,146,225,180,73,14,1,9,64,231,456,501,304,99,16,1,10,81,344,833,
%U A104698 1182,985,476,129,18,1,11,100,489,1408,2471,2668,1765,704,163,20,1,12
%N A104698 Triangle read by rows: T(n,k) = Sum_{j=0..n-k} binomial(k, j)*binomial(n-j+1, k+1).
%C A104698 The n-th column of the triangle is the binomial transform of the n-th row of A081277, followed by zeros. Example: column 3, (1, 6, 19, 44, ...) = binomial transform of row 3 of A081277: (1, 5, 8, 4, 0, 0, 0, ...). A104698 = reversal by rows of A142978. - _Gary W. Adamson_, Jul 17 2008
%C A104698 This sequence is jointly generated with A210222 as an array of coefficients of polynomials u(n,x): initially, u(1,x)=v(1,x)=1; for n > 1, u(n,x) = x*u(n-1,x) + v(n-1) + 1 and v(n,x) = 2x*u(n-1,x) + v(n-1,x) + 1. See the Mathematica section at A210222. - _Clark Kimberling_, Mar 19 2012
%C A104698 This Riordan triangle T appears in a formula for A001100(n, 0) = A002464(n), for n >= 1. - _Wolfdieter Lang_, May 13 2025
%H A104698 Reinhard Zumkeller, <a href="/A104698/b104698.txt">First 100 rows of triangle, flattened</a>
%F A104698 The triangle is extracted from the product A * B; A = [1; 1, 1; 1, 1, 1; ...], B = [1; 1, 1; 1, 3, 1; 1, 5, 5, 1; ...] both infinite lower triangular matrices (rest of the terms are zeros). The triangle of matrix B by rows = A008288, Delannoy numbers.
%F A104698 From _Paul Barry_, Jul 18 2005: (Start)
%F A104698 Riordan array (1/(1-x)^2, x(1+x)/(1-x)) = (1/(1-x), x)*(1/(1-x), x(1+x)/(1-x)).
%F A104698 T(n, k) = Sum_{j=0..n} Sum_{i=0..j-k} C(j-k, i)*C(k, i)*2^i.
%F A104698 T(n, k) = Sum_{j=0..k} Sum_{i=0..n-k-j} (n-k-j-i+1)*C(k, j)*C(k+i-1, i). (End)
%F A104698 T(n, k) = binomial(n+1, k+1)*2F1([-k, k-n], [-n-1], -1) where 2F1 is a Gaussian hypergeometric function. - _R. J. Mathar_, Sep 04 2011
%F A104698 T(n, k) = T(n-2, k-1) + T(n-1, k-1) + T(n-1, k) for 1 < k < n; T(n, 0) = n + 1; T(n, n) = 1. - _Reinhard Zumkeller_, Jul 17 2015
%F A104698 From _Wolfdieter Lang_, May 13 2025: (Start)
%F A104698 The Riordan triangle T = (1/(1 - x)^2, x*(1 + x)/(1 - x)) has the o.g.f. G(x, y) = 1/((1 - x)*(1 - x - y*x*(1+x))) for the row polynomials R(n, y) = Sum_{k=0..n} T(n, k)*y^k.
%F A104698 The o.g.f. for column k is G(k, x) = (1/(1 - x)^2)*(x*(1 + x)/(1 - x))^k, for k >= 0.
%F A104698 The o.g.f. for the diagonal m is D(m, x) = N(m, x)/(1 - x)^(m+1), with the numerator polynomial N(m, x) = Sum_{k=0..floor(m/2)} A034867(m, k)*x^(2*k) for m >= 0.
%F A104698 The row sums with o.g.f. R(x) = 1/((1 -x)*(1 - 2*x -x^2) give A048739.
%F A104698 The alternating row sums with o.g.f. 1/((1 - x)(1 + x^2)) give A133872.
%F A104698 The A-sequence for this Riordan triangle has o.g.f. A(x) = 1 + x + sqrt(1 + 6*x + x^2))/2 giving A112478(n). Hence T(n, k) = Sum_{j=0..n-k} A112478(j)*T(n-1, k-1+j), for n >= 1, k >= 1, T(n, k) = 0 for n < k, and T(0, 0) = 1.
%F A104698 The Z-sequence has o.g.f. (5 + x - sqrt(1 + 6*x + x^2))/2 = 3 + x - A(x) giving Z(n) = {2, -1, -A112478(n >= 2)}. Hence T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), for n >= 1. For A- and Z-sequences of Riordan triangles see a W. Lang link at A006232 with references.
%F A104698 The Boas-Buck sequences alpha and beta for the Riordan triangle T (see A046521 for the Aug 10 2017 comment and reference) are alpha(n) = A040000(n+1) = repeat{2} and beta(n) = A010673(n+1) = repeat{2,0}. Hence the recurrence for column T(n, k)_{n>=k}, with input T(k, k) = 1, for k >= 0, is T(n, k) = (1/(n-k)) * Sum_{j=k..n-1} (2 + k*(1 + (-1)^(n-1-j))) *T(j,k), for n >= k+1. (End)
%e A104698 The Riordan triangle T begins:
%e A104698 n\k 0 1 2 3 4 5 6 7 8 9 10 ...
%e A104698 ----------------------------------------------------
%e A104698 0: 1
%e A104698 1: 2 1
%e A104698 2: 3 4 1
%e A104698 3: 4 9 6 1
%e A104698 4: 5 16 19 8 1
%e A104698 5: 6 25 44 33 10 1
%e A104698 6: 7 36 85 96 51 12 1
%e A104698 7: 8 49 146 225 180 73 14 1
%e A104698 8: 9 64 231 456 501 304 99 16 1
%e A104698 9: 10 81 344 833 1182 985 476 129 18 1
%e A104698 10: 11 100 489 1408 2471 2668 1765 704 163 20 1
%e A104698 ... reformatted and extended by _Wolfdieter Lang_, May 13 2025
%e A104698 From _Wolfdieter Lang_, May 13 2025: (Start)
%e A104698 Zumkeller recurrence (adapted for offset [0,0]): 19 = T(4, 2) = T(2, 1) + T(3, 1) + T(3,3) = 4 + 9 + 6 = 19.
%e A104698 A-sequence recurrence: 19 = T(4, 2) = 1*T(3. 1) + 2*T(3. 2) - 2*T(3, 3) = 9 + 12 - 2 = 19.
%e A104698 Z-sequence recurrence: 5 = T(4, 0) = 2*T(3, 0) - 1*T(3, 1) + 2*T(3, 2) - 6*T(3, 3) = 8 - 9 + 12 + 6 = 5.
%e A104698 Boas-Buck recurrence: 19 = T(4, 2) = (1/2)*((2 + 0)*T(2, 2) + (2 + 2*2)*T(3, 2)) = (1/2)*(2 + 36) = 19. (End)
%p A104698 A104698 := proc(n, k) add(binomial(k, j)*binomial(n-j+1, n-k-j), j=0..n-k) ; end proc:
%p A104698 seq(seq(A104698(n, k), k=0..n), n=0..15); # _R. J. Mathar_, Sep 04 2011
%p A104698 T := (n, k) -> binomial(n + 1, k + 1)*hypergeom([-k, k - n], [-n - 1], -1):
%p A104698 for n from 0 to 9 do seq(simplify(T(n, k)), k = 0..n) od;
%p A104698 T := proc(n, k) option remember; if k = 0 then n + 1 elif k = n then 1 else T(n-2, k-1) + T(n-1, k-1) + T(n-1, k) fi end: # _Peter Luschny_, May 13 2025
%t A104698 u[1, _] = 1; v[1, _] = 1;
%t A104698 u[n_, x_] := u[n, x] = x u[n-1, x] + v[n-1, x] + 1;
%t A104698 v[n_, x_] := v[n, x] = 2 x u[n-1, x] + v[n-1, x] + 1;
%t A104698 Table[CoefficientList[u[n, x], x], {n, 1, 11}] // Flatten (* _Jean-François Alcover_, Mar 10 2019, after _Clark Kimberling_ *)
%o A104698 (PARI) T(n,k)=sum(j=0,n-k,binomial(k,j)*binomial(n-j+1,k+1)) \\ _Charles R Greathouse IV_, Jan 16 2012
%o A104698 (Haskell)
%o A104698 a104698 n k = a104698_tabl !! (n-1) !! (k-1)
%o A104698 a104698_row n = a104698_tabl !! (n-1)
%o A104698 a104698_tabl = [1] : [2,1] : f [1] [2,1] where
%o A104698 f us vs = ws : f vs ws where
%o A104698 ws = zipWith (+) ([0] ++ us ++ [0]) $
%o A104698 zipWith (+) ([1] ++ vs) (vs ++ [0])
%o A104698 -- _Reinhard Zumkeller_, Jul 17 2015
%Y A104698 Diagonal sums are A008937(n+1).
%Y A104698 Cf. A048739 (row sums), A008288, A005900 (column 3), A014820 (column 4)
%Y A104698 Cf. A081277, A142978 by antidiagonals, A119328, A110271 (matrix inverse).
%Y A104698 Cf. A001100, A002464, A010673, A040000, A112478, A133872, A383857.
%K A104698 nonn,easy,tabl
%O A104698 0,2
%A A104698 _Gary W. Adamson_, Mar 19 2005