Werner Schulte - cp's OEIS frontend

A385590 Triangle read by rows, based on Fibonacci numbers: Let i > 1 be such that F(i) <= n < F(i+1); i.e., i = A130233(n). Then T(n, k) = F(i-1)^2 + 1 - (i-1) mod 2 + (n - F(i)) * F(i-2) + (k-1) * F(i-1) where F(k) = A000045(k).

1, 2, 3, 4, 6, 8, 5, 7, 9, 11, 10, 13, 16, 19, 22, 12, 15, 18, 21, 24, 27, 14, 17, 20, 23, 26, 29, 32, 25, 30, 35, 40, 45, 50, 55, 60, 28, 33, 38, 43, 48, 53, 58, 63, 68, 31, 36, 41, 46, 51, 56, 61, 66, 71, 76, 34, 39, 44, 49, 54, 59, 64, 69, 74, 79, 84, 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, 65, 73, 81, 89, 97

Offset: 1

Werner Schulte, Jul 03 2025

Conjecture: This triangle yields a permutation of the natural numbers.

			Triangle T(n, k) for 1 <= k <= n starts:
n\ k :   1   2   3   4   5   6   7   8   9  10  11  12  13
==========================================================
   1 :   1
   2 :   2   3
   3 :   4   6   8
   4 :   5   7   9  11
   5 :  10  13  16  19  22
   6 :  12  15  18  21  24  27
   7 :  14  17  20  23  26  29  32
   8 :  25  30  35  40  45  50  55  60
   9 :  28  33  38  43  48  53  58  63  68
  10 :  31  36  41  46  51  56  61  66  71  76
  11 :  34  39  44  49  54  59  64  69  74  79  84
  12 :  37  42  47  52  57  62  67  72  77  82  87  92
  13 :  65  73  81  89  97 105 113 121 129 137 145 153 161
  etc.

Cf. A000045, A130233.

T(n, k) = i=1; for(j=1,n,if(j==fibonacci(i+1),i=i+1)); (fibonacci(i-1))^2+1-(i-1)%2 + (n-fibonacci(i))*fibonacci(i-2) + (k-1)*fibonacci(i-1)

Conjecture: Sum_{k=1..n} (-1)^k * binomial(n-1, k-1) * T(n, k) = 0 for n > 2 and (-1)^n for n < 3.

A385406 Triangle read by rows: T(n, k) = n(n+1)/2 - floor((n-1)/2) - (-1)^k floor(k/2).

Original entry on oeis.org

1, 3, 2, 5, 4, 6, 9, 8, 10, 7, 13, 12, 14, 11, 15, 19, 18, 20, 17, 21, 16, 25, 24, 26, 23, 27, 22, 28, 33, 32, 34, 31, 35, 30, 36, 29, 41, 40, 42, 39, 43, 38, 44, 37, 45, 51, 50, 52, 49, 53, 48, 54, 47, 55, 46, 61, 60, 62, 59, 63, 58, 64, 57, 65, 56, 66, 73, 72, 74, 71, 75, 70, 76, 69, 77, 68, 78, 67

Offset: 1

Werner Schulte, Jun 27 2025

This triangle seen as a sequence yields a permutation of the natural numbers (A000027).

			Triangle T(n, k) for 1 <= k <= n starts:
n \k :   1   2   3   4   5   6   7   8   9  10  11  12  13
==========================================================
   1 :   1
   2 :   3   2
   3 :   5   4   6
   4 :   9   8  10   7
   5 :  13  12  14  11  15
   6 :  19  18  20  17  21  16
   7 :  25  24  26  23  27  22  28
   8 :  33  32  34  31  35  30  36  29
   9 :  41  40  42  39  43  38  44  37  45
  10 :  51  50  52  49  53  48  54  47  55  46
  11 :  61  60  62  59  63  58  64  57  65  56  66
  12 :  73  72  74  71  75  70  76  69  77  68  78  67
  13 :  85  84  86  83  87  82  88  81  89  80  90  79  91
  etc.

Index entries for sequences that are permutations of the natural numbers

Cf. A080827 (column 1), A128918 (main diagonal), A006003 (row sums), A213399.

T[n_, k_] := n*(n+1)/2 - Floor[(n-1)/2] - (-1)^k*Floor[k/2]; Table[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Amiram Eldar, Jun 28 2025 *)

T(n, k) = n*(n+1)/2 - floor((n-1)/2) - (-1)^k * floor(k/2)

T(n, k) = T(n, k-1) - (-1)^k * (k-1) for 1 < k <= n with initial values T(n, 1) = n*(n+1)/2 - floor((n-1)/2) for n >= 1.
T(n, n) = n*(n+1)/2 + (1-n) * (1 - n mod 2) = A128918(n).
T(2*n-1, n) = 2*n^2 - 2*n + 1 - (-1)^n * floor(n/2) = A213399(n-1).

A384531 Multiplicative sequence a(n) with a(p^e) = ((2e+1) p - 2e) p^(e-1) for prime p and e >= 0.

Original entry on oeis.org

1, 4, 7, 12, 13, 28, 19, 32, 33, 52, 31, 84, 37, 76, 91, 80, 49, 132, 55, 156, 133, 124, 67, 224, 105, 148, 135, 228, 85, 364, 91, 192, 217, 196, 247, 396, 109, 220, 259, 416, 121, 532, 127, 372, 429, 268, 139, 560, 217, 420, 343, 444, 157, 540, 403, 608, 385, 340, 175, 1092

Offset: 1

Werner Schulte, Jun 01 2025

Vaclav Kotesovec, Graph - the asymptotic ratio (100000 terms)
David Radcliffe, Notes on two arithmetical functions (Contains a proof that a(n) = Sum_{i=1..n} gcd(i, n) * gcd(i+1, n)).

Cf. A018804, A173557.

Cf. A065473.

A384531 := proc(n)
    local a,pe,p,e;
    a :=1 ;
    for pe in ifactors(n)[2] do
        p := op(1,pe) ;
        e := op(2,pe) ;
        a := a*((2*e+1) * p - 2*e) * p^(e-1) ;
    end do:
    a ;
end proc:
seq(A384531(n),n=1..100) ;# R. J. Mathar, Jun 04 2025

f[p_, e_] := ((2*e+1)*p - 2*e)*p^(e-1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Jun 01 2025 *)

a(n)=my(f=factor(n)); prod(k=1,#f[,1],((2*f[k,2]+1)*f[k,1]-2*f[k,2])*f[k,1]^(f[k,2]-1))

from math import prod
from sympy import factorint
def A384531(n): return prod((((m:=e<<1)|1)*p-m)*p**(e-1) for p, e in factorint(n).items()) # Chai Wah Wu, Jun 03 2025

Dirichlet g.f.: Sum_{n > 0} a(n) / n^s = (zeta(s-1))^2 * Product_{p prime} (1 + (p-2) / p^s).
Dirichlet convolution of A018804 and A173557.
Conjecture: a(n) = Sum_{i=1..n} gcd(i, n) * gcd(i+1, n).
From Vaclav Kotesovec, Jun 04 2025: (Start)
Let f(s) = Product_{primes p} (1 + 2/p^(2*s-1) - 1/p^(2*s-2) - 2/p^s).
Dirichlet g.f.: zeta(s-1)^3 * f(s).
Sum_{k=1..n} a(k) ~ f(2) * n^2 * (log(n)^2 + (6*gamma - 1 + 2*f'(2)/f(2))*log(n) + 1/2 - 3*gamma + 6*gamma^2 - 6*sg1 + (6*gamma - 1)*f'(2)/f(2) + f''(2)/f(2))/4, where
f(2) = Product_{primes p} (1 - 3/p^2 + 2/p^3) = A065473 = 0.2867474284344787341...,
f'(2) = f(2) * Sum_{primes p} 4*log(p)/(p^2 + p - 2) = 0.53488225650873164189786660885838556843579696135554271633442328...,
f''(2) = f'(2)^2/f(2) + f(2) * Sum_{primes p} (-2*p*(3*p+2)*log(p)^2 / (p^2+p-2)^2) = -0.29112624105319980992840485620511000074444413707069816872854442...,
gamma is the Euler-Mascheroni constant A001620 and sg1 is the first Stieltjes constant (see A082633). (End)

A380191 Triangle read by rows: Riordan array (2 - D(x), x * D(x)) where D(x) is g.f. of A001764.

Original entry on oeis.org

1, -1, 1, -3, 0, 1, -12, -1, 1, 1, -55, -6, 2, 2, 1, -273, -33, 5, 6, 3, 1, -1428, -182, 13, 22, 11, 4, 1, -7752, -1020, 28, 91, 46, 17, 5, 1, -43263, -5814, 0, 408, 210, 78, 24, 6, 1, -246675, -33649, -627, 1938, 1020, 380, 119, 32, 7, 1, -1430715, -197340, -6325, 9614, 5187, 1938, 612, 170, 41, 8, 1

Offset: 0

Werner Schulte, Jan 15 2025

			Triangle T(n, k) for 0 <= k <= n starts:
n \k :         0        1      2     3     4     5    6    7   8  9  10
=======================================================================
   0 :         1
   1 :        -1        1
   2 :        -3        0      1
   3 :       -12       -1      1     1
   4 :       -55       -6      2     2     1
   5 :      -273      -33      5     6     3     1
   6 :     -1428     -182     13    22    11     4    1
   7 :     -7752    -1020     28    91    46    17    5    1
   8 :    -43263    -5814      0   408   210    78   24    6   1
   9 :   -246675   -33649   -627  1938  1020   380  119   32   7  1
  10 :  -1430715  -197340  -6325  9614  5187  1938  612  170  41  8   1
  etc.

Cf. A001764, A110616.

T(n, k) = if(k==n, 1, binomial(3*n-2*k, n-k) * (n*k+4*k-3*n) / ((3*n-2*k) * (2*n-k+1)))

T(n, k) = binomial(3*n - 2*k, n - k) * (n*k + 4*k - 3*n) / ((3*n - 2*k) * (2*n - k + 1)) if 0 <= k < n, and T(n, n) = 1 for n >= 0.
G.f.: (2 - D(t)) / (1 - x * t * D(t)) where D(t) is g.f. of A001764.
Conjecture: Sum_{i=0..n-k} binomial(2*i, i) * T(n, i+k) = A110616(n, k).

A380128 Triangle read by rows: Riordan array (1/(C(x)sqrt(1-4x)), x/C(x)) where C(x) is g.f. of A000108.

Original entry on oeis.org

1, 1, 1, 3, 0, 1, 10, 1, -1, 1, 35, 4, 0, -2, 1, 126, 15, 1, 0, -3, 1, 462, 56, 5, 0, 1, -4, 1, 1716, 210, 21, 1, 0, 3, -5, 1, 6435, 792, 84, 6, 0, 0, 6, -6, 1, 24310, 3003, 330, 28, 1, 0, -1, 10, -7, 1, 92378, 11440, 1287, 120, 7, 0, 0, -4, 15, -8, 1, 352716, 43758, 5005, 495, 36, 1, 0, 0, -10, 21, -9, 1

Offset: 0

Werner Schulte, Jan 12 2025

			Triangle T(n, k) for 0 <= k <= n starts:
n \k :       0      1     2    3   4   5   6   7    8   9  10  11
=================================================================
   0 :       1
   1 :       1      1
   2 :       3      0     1
   3 :      10      1    -1    1
   4 :      35      4     0   -2   1
   5 :     126     15     1    0  -3   1
   6 :     462     56     5    0   1  -4   1
   7 :    1716    210    21    1   0   3  -5   1
   8 :    6435    792    84    6   0   0   6  -6    1
   9 :   24310   3003   330   28   1   0  -1  10   -7   1
  10 :   92378  11440  1287  120   7   0   0  -4   15  -8   1
  11 :  352716  43758  5005  495  36   1   0   0  -10  21  -9   1
  etc.

Cf. A088218 (column 0), A001791 (column 1), A000007 (central terms).

Cf. A000108, A001764.

T(n, k) = polcoef(polcoef(1/((C(t)-x*t)*sqrt(1-4*t))+x*O(x^k), k, x)+t*O(t^n), n, t);
       m = matrix(15, 15, n, k, if(k>n, 0, T(n-1, k-1)))

G.f.: 1/((C(t) - x*t) * sqrt(1 - 4*t)) where C(t) is g.f. of A000108.
Inverse Riordan array is (2 - D(x), x * D(x)) where D(x) is g.f. of A001764.
Conjecture: T(n, k) = Sum_{i=0..n-k} (-1)^i * binomial(2*n - 2*k, n - k - i) * binomial(k + i, i).

A379907 Triangle read by rows: T(n, k) = Sum_{i=0..n-k} (-1)^(n - k - i) * binomial(n - k, i) * binomial(k + 2i, i) (k + 1) / (k + 1 + i).

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 1, 2, 2, 1, 3, 4, 4, 3, 1, 6, 9, 9, 7, 4, 1, 15, 21, 21, 17, 11, 5, 1, 36, 51, 51, 42, 29, 16, 6, 1, 91, 127, 127, 106, 76, 46, 22, 7, 1, 232, 323, 323, 272, 200, 128, 69, 29, 8, 1, 603, 835, 835, 708, 530, 352, 204, 99, 37, 9, 1, 1585, 2188, 2188, 1865, 1415, 965, 587, 311, 137, 46, 10, 1

Offset: 0

Werner Schulte, Jan 05 2025

Conjecture: Let A = (g(t), f(t)) and B = (u(t), v(t)) be (triangular) Riordan arrays with A(n, k) = [t^n](g(t)*(f(t))^k) and B(n, k) = [t^n](u(t)*(v(t))^k). Then T = (g(t)*u(f(t)), v(f(t))*t/f(t)) is the Riordan array with T(n, k) = [t^n](g(t)*u(f(t))*(v(f(t))*t/f(t))^k) = Sum_{i=0..n-k} A(n-k, i) * B(k+i, k) for 0 <= k <= n.

			Triangle T(n, k) for 0 <= k <= n starts:
n \k :     0     1     2     3     4    5    6    7    8   9  10  11
====================================================================
   0 :     1
   1 :     0     1
   2 :     1     1     1
   3 :     1     2     2     1
   4 :     3     4     4     3     1
   5 :     6     9     9     7     4    1
   6 :    15    21    21    17    11    5    1
   7 :    36    51    51    42    29   16    6    1
   8 :    91   127   127   106    76   46   22    7    1
   9 :   232   323   323   272   200  128   69   29    8   1
  10 :   603   835   835   708   530  352  204   99   37   9   1
  11 :  1585  2188  2188  1865  1415  965  587  311  137  46  10   1
  etc.

Cf. A005043 (column 0), A001006 (column 1 and 2), A102071 (column 3).

Cf. A000108, A342912 (row sums), A379824 (alternating row sums), A379823 (central terms).

gf := 2/(sqrt((1-3*t)*(t+1)) - 2*(t+1)*t*x + t+1): ser := simplify(series(gf,t,12)):
ct := n -> coeff(ser,t,n): row := n -> local k; seq(coeff(ct(n), x, k), k = 0..n):
seq(row(n), n = 0..11);  # Peter Luschny, Jan 05 2025

T(n,k) = sum(i=0,n-k,(-1)^(n-k-i)*binomial(n-k,i)*binomial(k+2*i,i)*(k+1)/(k+1+i))

T(n,k)=polcoef(polcoef(2/(sqrt((1-3*t)*(1+t))+(1+t)*(1-2*x*t))+x*O(x^k),k,x)+t*O(t^n),n,t);
       m=matrix(15,15,n,k,if(k>n,0,T(n-1,k-1)))

Riordan array (C(t/(1+t)) / (1+t), t * C(t/(1+t))) where C(x) is g.f. of A000108.
Riordan array ((1 + t - sqrt(1 - 2*t - 3*t^2))/(2*t*(1 + t)), (1 + t - sqrt(1-2*t-3*t^2))/2).
G.f.: 2/(sqrt((1 - 3*t)*(t + 1)) - 2*(t + 1)*t*x + t + 1).
Conjecture: T(n, k) = T(n, k-1) + T(n-1, k-1) - T(n-1, k-2) - T(n-2, k-2) for 2 <= k <= n.
T(n, k) = (-1)^(k-n)*hypergeom([k-n, k/2+1, (k+1)/2], [1, k + 2], 4). - Peter Luschny, Jan 06 2025

A378931 Triangle read by rows, based on products of Jacobsthal numbers (A001045).

Original entry on oeis.org

1, -1, 3, -2, -9, 15, -4, -18, -25, 55, -8, -36, -50, -121, 231, -16, -72, -100, -242, -441, 903, -32, -144, -200, -484, -882, -1849, 3655, -64, -288, -400, -968, -1764, -3698, -7225, 14535, -128, -576, -800, -1936, -3528, -7396, -14450, -29241, 58311, -256, -1152, -1600, -3872, -7056, -14792, -28900, -58482, -116281, 232903

Offset: 1

Werner Schulte, Dec 11 2024

Let M = T^(-1) be matrix inverse of T seen as a lower triangular matrix. M is a harmonic triangle with M(n, k) = 1 / A084175(n) if k = n, and 1 / A378676(k) if 1 <= k < n. Triangle M(n, k) for 1 <= k <= n starts:
  1/1
  1/3  1/3
  1/3  1/5  1/15
  1/3  1/5  1/33   1/55
  1/3  1/5  1/33  1/105  1/231
  1/3  1/5  1/33  1/105  1/473  1/903
  etc.
  Sum_{k=1..n} M(n, k) * 2^(k-1) = 1.
  Sum_{k=1..n} M(n, k) * (-2)^(k-1) = (-1)^(n-1) / A001045(n+1).
  Sum_{k=1..n} 2^(k-1) / M(n, k) = (8^n - 1) / 7 = A023001(n).

			Triangle T(n, k) for 1 <= k <= n starts:
n\k :     1     2     3      4      5      6       7       8      9
===================================================================
  1 :     1
  2 :    -1     3
  3 :    -2    -9    15
  4 :    -4   -18   -25     55
  5 :    -8   -36   -50   -121    231
  6 :   -16   -72  -100   -242   -441    903
  7 :   -32  -144  -200   -484   -882  -1849    3655
  8 :   -64  -288  -400   -968  -1764  -3698   -7225   14535
  9 :  -128  -576  -800  -1936  -3528  -7396  -14450  -29241  58311
  etc.

Cf. A001045, A023001, A378676.

A084175 (main diagonal), A139818 (1st subdiagonal), A000079 (column 1 and row sums).

T[n_,k_]:=If[k==n, (2*4^n-(-2)^n-1)/9, -2^(n-1-k)*(2^(k+1)+(-1)^k)^2/9]; Table[T[n,k],{n,10},{k,n}]//Flatten (* Stefano Spezia, Dec 11 2024 *)

T(n,k)=if(k==n,(2*4^n-(-2)^n-1)/9,-2^(n-1-k)*(2^(k+1)+(-1)^k)^2/9)

T(n, n) = (2 * 4^n - (-2)^n - 1) / 9 = A084175(n), and T(n, k) = -2^(n-1-k) * (2^(k+1) + (-1)^k)^2 / 9 for 1 <= k < n.

G.f.: x*t * (1 - 3*t - 6*x*t^2 + 8*x^2*t^3) / ((1 - 2*t) * (1 - x*t) * (1 + 2*x*t) * (1 - 4*x*t)).

A378676 a(n) = J(n) * J(n+2) where J(n) = Jacobsthal(n) = A001045(n).

Original entry on oeis.org

0, 3, 5, 33, 105, 473, 1785, 7353, 28985, 116793, 465465, 1865273, 7454265, 29830713, 119295545, 477236793, 1908837945, 7635570233, 30541844025, 122168249913, 488671252025, 1954688503353, 7818747022905, 31275002072633, 125099980328505, 500399977238073, 2001599797104185, 8006399412112953, 32025597201059385

Offset: 0

Werner Schulte, Dec 03 2024

Index entries for linear recurrences with constant coefficients, signature (3,6,-8).

Cf. A001045, A084175.

a[n_] := (4^(n+1) - 5*(-2)^n + 1)/9; Array[a, 30, 0] (* Amiram Eldar, Dec 06 2024 *)

PARI
```
a(n)=(4^(n+1)-5*(-2)^n+1)/9
```

a(n) = (2^n - (-1)^n) * (2^(n+2) - (-1)^n) / 9 = (4 * 4^n - 5 * (-2)^n + 1) / 9.
G.f.: x * (3 - 4*x) / ((1-x) * (1+2*x) * (1-4*x)).
a(n) = 3 * a(n-1) + 6 * a(n-2) - 8 * a(n-3) for n > 2 with initial values a(0) = 0, a(1) = 3, and a(2) = 5.
Sum_{k=1..n-1} 2^(k-1) / a(k) = 1 - 2^(n-1) / A084175(n) for n > 0.
Sum_{k>0} 2^(k-1) / a(k) = 1.
E.g.f.: exp(x)*(1 - cosh(3*x) + 9*sinh(3*x))/9. - Stefano Spezia, Dec 06 2024

A378277 Denominators in a harmonic triangle, based on products of Fibonacci numbers.

Original entry on oeis.org

1, 2, 2, 2, 3, 6, 2, 3, 10, 15, 2, 3, 10, 24, 40, 2, 3, 10, 24, 65, 104, 2, 3, 10, 24, 65, 168, 273, 2, 3, 10, 24, 65, 168, 442, 714, 2, 3, 10, 24, 65, 168, 442, 1155, 1870, 2, 3, 10, 24, 65, 168, 442, 1155, 3026, 4895, 2, 3, 10, 24, 65, 168, 442, 1155, 3026, 7920, 12816

Offset: 1

Werner Schulte, Nov 21 2024

The harmonic triangle uses the terms of this sequence as denominators, numerators = 1.

The inverse of the harmonic triangle has entries -(Fibonacci(k+1))^2 for 1<=k

Row sums of the harmonic triangle are 1.

Conjecture: Alt. row sums of the harmonic triangle are Fibonacci(n-2) / Fibonacci(n+1), where Fibonacci(-1) = 1.

			Triangle T(n, k) for 1 <= k <= n starts:
n\ k :  1  2   3   4   5    6    7     8     9    10     11
===========================================================
   1 :  1
   2 :  2  2
   3 :  2  3   6
   4 :  2  3  10  15
   5 :  2  3  10  24  40
   6 :  2  3  10  24  65  104
   7 :  2  3  10  24  65  168  273
   8 :  2  3  10  24  65  168  442   714
   9 :  2  3  10  24  65  168  442  1155  1870
  10 :  2  3  10  24  65  168  442  1155  3026  4895
  11 :  2  3  10  24  65  168  442  1155  3026  7920  12816
  etc.

Cf. A000045, A110034, A110035, A001654 (main diagonal), A059929 (subdiagonals).

T(n,k)=if(k==n,Fibonacci(n)*Fibonacci(n+1),Fibonacci(k)*Fibonacci(k+2))

T(n, k) = Fibonacci(n) * Fibonacci(n+1) if k = n, and Fibonacci(k) * Fibonacci(k+2) if 1 <= k < n.
Row sums are A110035(n) - 1 = -A110034(n+1).
G.f.: A(t, x) = x*t*(1 + t - x*t^2) / ((1 - t) * (1 + x*t) * (1 - 3*x*t + x^2*t^2)).

A378145 Riordan triangle (1 + x * C(x), x * C(x)), where C(x) is g.f. of A000108.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 4, 3, 1, 5, 10, 8, 4, 1, 14, 28, 23, 13, 5, 1, 42, 84, 70, 42, 19, 6, 1, 132, 264, 222, 138, 68, 26, 7, 1, 429, 858, 726, 462, 240, 102, 34, 8, 1, 1430, 2860, 2431, 1573, 847, 385, 145, 43, 9, 1, 4862, 9724, 8294, 5434, 3003, 1430, 583, 198, 53, 10, 1

Offset: 0

Werner Schulte, Nov 17 2024

			Triangle T(n, k) for 0 <= k <= n starts:
n\k :     0     1     2     3    4    5    6   7  8  9
======================================================
  0 :     1
  1 :     1     1
  2 :     1     2     1
  3 :     2     4     3     1
  4 :     5    10     8     4    1
  5 :    14    28    23    13    5    1
  6 :    42    84    70    42   19    6    1
  7 :   132   264   222   138   68   26    7   1
  8 :   429   858   726   462  240  102   34   8  1
  9 :  1430  2860  2431  1573  847  385  145  43  9  1
  etc.

Cf. A000108, A004070, A120588 (column 0), A068875 (column 1 and row sums), A000007 (alt. row sums).

T(n,k)=if(k==n,1,binomial(2*n-k,n)*(n*(3*k+1)-2*k*(k+1))/((2*n-k)*(2*n-k-1)))

T(n, k) = binomial(2*n-k, n) * (n*(3*k+1) - 2*k*(k+1)) / ((2*n-k) * (2*n-k-1)) if 0 <= k < n and 1 if k = n.
T(n, k) = T(n, k-1) - T(n-1, k-2) for 2 <= k <= n.
(-1)^(n-k) * T(n, k) is matrix inverse of A004070 (seen as a triangle).
Conjecture: Sum_{i=0..n-k} binomial(i+m-1, i) * T(n, i+k) = T(n+m, m+k) for m > 0.
Conjecture: Sum_{k=0..n} (1 + floor(k/2)) * T(n, k) = A000108(n+1).
G.f.: A(x, y) = (1 + x*C(x)) / (1 - y * x*C(x)), where C(x) is g.f. of A000108.

cp's OEIS Frontend

User: Werner Schulte

A385590 Triangle read by rows, based on Fibonacci numbers: Let i > 1 be such that F(i) <= n < F(i+1); i.e., i = A130233(n). Then T(n, k) = F(i-1)^2 + 1 - (i-1) mod 2 + (n - F(i)) * F(i-2) + (k-1) * F(i-1) where F(k) = A000045(k).

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A385406 Triangle read by rows: T(n, k) = n*(n+1)/2 - floor((n-1)/2) - (-1)^k * floor(k/2).

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A384531 Multiplicative sequence a(n) with a(p^e) = ((2*e+1) * p - 2*e) * p^(e-1) for prime p and e >= 0.

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A380191 Triangle read by rows: Riordan array (2 - D(x), x * D(x)) where D(x) is g.f. of A001764.

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A380128 Triangle read by rows: Riordan array (1/(C(x)*sqrt(1-4*x)), x/C(x)) where C(x) is g.f. of A000108.

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A379907 Triangle read by rows: T(n, k) = Sum_{i=0..n-k} (-1)^(n - k - i) * binomial(n - k, i) * binomial(k + 2*i, i) * (k + 1) / (k + 1 + i).

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A378931 Triangle read by rows, based on products of Jacobsthal numbers (A001045).

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A378676 a(n) = J(n) * J(n+2) where J(n) = Jacobsthal(n) = A001045(n).

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A385406 Triangle read by rows: T(n, k) = n(n+1)/2 - floor((n-1)/2) - (-1)^k floor(k/2).

A384531 Multiplicative sequence a(n) with a(p^e) = ((2e+1) p - 2e) p^(e-1) for prime p and e >= 0.

A380128 Triangle read by rows: Riordan array (1/(C(x)sqrt(1-4x)), x/C(x)) where C(x) is g.f. of A000108.

A379907 Triangle read by rows: T(n, k) = Sum_{i=0..n-k} (-1)^(n - k - i) * binomial(n - k, i) * binomial(k + 2i, i) (k + 1) / (k + 1 + i).