A026010 -id:A026010 - cp's OEIS frontend

A325699 Number of distinct even prime indices of n minus the number of distinct odd prime indices of n.

0, -1, 1, -1, -1, 0, 1, -1, 1, -2, -1, 0, 1, 0, 0, -1, -1, 0, 1, -2, 2, -2, -1, 0, -1, 0, 1, 0, 1, -1, -1, -1, 0, -2, 0, 0, 1, 0, 2, -2, -1, 1, 1, -2, 0, -2, -1, 0, 1, -2, 0, 0, 1, 0, -2, 0, 2, 0, -1, -1, 1, -2, 2, -1, 0, -1, -1, -2, 0, -1, 1, 0, -1, 0, 0, 0

Offset: 1

Gus Wiseman, May 17 2019

sign

A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.

a(n) = A324967(n) - A324966(n).

Cf. A000712, A001221, A026010, A045931, A063886, A112798, A130780, A171966, A241638, A325698, A325700.

Table[Total[(-1)^PrimePi/@First/@If[n==1,{},FactorInteger[n]]],{n,100}]

G.f.: Sum_{k>=1} (-1)^k * x^prime(k) / (1 - x^prime(k)). - Ilya Gutkovskiy, Feb 12 2020

Additive with a(p^e) = (-1)^primepi(p). - Amiram Eldar, Jun 17 2024

A368175 Square array read by ascending antidiagonals: T(n,k) = Sum_{i=ceiling((k-n)/2)..floor((k+n-1)/2)} binomial(k,i), with n >= 1, k >= 0.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 2, 3, 3, 1, 2, 4, 6, 6, 1, 2, 4, 7, 10, 10, 1, 2, 4, 8, 14, 20, 20, 1, 2, 4, 8, 15, 25, 35, 35, 1, 2, 4, 8, 16, 30, 50, 70, 70, 1, 2, 4, 8, 16, 31, 56, 91, 126, 126, 1, 2, 4, 8, 16, 32, 62, 112, 182, 252, 252, 1, 2, 4, 8, 16, 32, 63, 119, 210, 336, 462, 462

Offset: 1

Paolo Xausa, Dec 14 2023

T(n,k), for k >= 1, is the size of the largest possible set S of k-bit strings such that, if S_a < S_b are members of S, then W(S_b) < W(S_a) + n, where W is A000120.

T(1,k), for k >= 1, gives the number of rows in the Christmas tree pattern (cf. A367508) of order k. Furthermore, T(n,k), for k >= 1, gives the number of rows generated by iteratively applying k times the map described in A367508, starting from a single row of length n.

			Array begins:
  n\k|  0  1  2  3   4   5   6    7    8    9    10  ...
  ---+--------------------------------------------------
   1 |  1, 1, 2, 3,  6, 10, 20,  35,  70, 126,  252, ... = A001405
   2 |  1, 2, 3, 6, 10, 20, 35,  70, 126, 252,  462, ... = A001405
   3 |  1, 2, 4, 7, 14, 25, 50,  91, 182, 336,  672, ... = A026010
   4 |  1, 2, 4, 8, 15, 30, 56, 112, 210, 420,  792, ... = A026023
   5 |  1, 2, 4, 8, 16, 31, 62, 119, 238, 456,  912, ...
   6 |  1, 2, 4, 8, 16, 32, 63, 126, 246, 492,  957, ...
   7 |  1, 2, 4, 8, 16, 32, 64, 127, 254, 501, 1002, ...
   8 |  1, 2, 4, 8, 16, 32, 64, 128, 255, 510, 1012, ...
   9 |  1, 2, 4, 8, 16, 32, 64, 128, 256, 511, 1022, ...
  10 |  1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1023, ...
  ...
For n = 3 and k = 4 the 14 members of S are 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, 1001, 1010, 1011, 1100, 1101, 1110.

Donald E. Knuth, The Art of Computer Programming, Vol. 4A: Combinatorial Algorithms, Part 1, Addison-Wesley, 2011, Section 7.2.1.6, exercises 71 and 72, pp. 479 and 799.

Paolo Xausa, Table of n, a(n) for n = 1..11325 (antidiagonals 1..150 of the array, flattened).

Cf. A000079, A000120, A000225, A001405, A014495, A026010, A026023, A191781, A367508.

A368175[n_,k_]:=If[n>k,2^k,Sum[Binomial[k,i],{i,Ceiling[(k-n)/2],Floor[(k+n-1)/2]}]];
With[{dmax=15},Table[A368175[n-k,k],{n,dmax},{k,0,n-1}]] (* Generates 15 antidiagonals *)

T(n,0) = 1.
T(1,k) = A001405(k).
T(n,k) = 2^k = A000079(k), for n > k.
T(n,n) = 2^n - 1 = A000225(n).
Antidiagonal sums: Sum_{n=1..d} T(n,d-n) = binomial(d+1,floor((d+1)/2)) - 1 = A014495(d+1), for d >= 1.

A118344 Pendular Catalan triangle, read by rows, where row n is formed from row n-1 by the recurrence: if n > 2k, T(n,k) = T(n,n-k) + T(n-1,k), else T(n,k) = T(n,n-1-k) - T(n-1,k) - T(n-1,k+1), for n>=k>=0, with T(n,0)=1 and T(n,n)=0^n.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 2, 1, 0, 1, 4, 5, 3, 1, 0, 1, 5, 9, 5, 4, 1, 0, 1, 6, 14, 14, 9, 5, 1, 0, 1, 7, 20, 28, 14, 14, 6, 1, 0, 1, 8, 27, 48, 42, 28, 20, 7, 1, 0, 1, 9, 35, 75, 90, 42, 48, 27, 8, 1, 0, 1, 10, 44, 110, 165, 132, 90, 75, 35, 9, 1, 0, 1, 11, 54, 154, 275, 297, 132, 165, 110, 44, 10, 1, 0

Offset: 0

Paul D. Hanna, Apr 26 2006

See A118340 for definition of pendular triangles and pendular sums.

			Row 6 equals the pendular sums of row 5:
  [1,  4,  5,  3,  1,  0], where the sums proceed as follows:
  [1, __, __, __, __, __]: T(6,0) = T(5,0) = 1;
  [1, __, __, __, __,  1]: T(6,5) = T(6,0) - T(5,5) = 1 - 0 = 1;
  [1,  5, __, __, __,  1]: T(6,1) = T(6,5) + T(5,1) = 1 + 4 = 5;
  [1,  5, __, __,  4,  1]: T(6,4) = T(6,1) - T(5,4) - T(5,5) = 5-1-0 = 4;
  [1,  5,  9, __,  4,  1]: T(6,2) = T(6,4) + T(5,2) = 4 + 5 = 9;
  [1,  5,  9,  5,  4,  1]: T(6,3) = T(6,2) - T(5,3) - T(5,4) = 9-3-1 = 5;
  [1,  5,  9,  5,  4,  1,  0] finally, append a zero to obtain row 6.
Triangle begins:
  1;
  1,  0;
  1,  1,  0;
  1,  2,  1,   0;
  1,  3,  2,   1,   0;
  1,  4,  5,   3,   1,   0;
  1,  5,  9,   5,   4,   1,   0;
  1,  6, 14,  14,   9,   5,   1,   0;
  1,  7, 20,  28,  14,  14,   6,   1,   0;
  1,  8, 27,  48,  42,  28,  20,   7,   1,  0;
  1,  9, 35,  75,  90,  42,  48,  27,   8,  1,  0;
  1, 10, 44, 110, 165, 132,  90,  75,  35,  9,  1,  0;
  1, 11, 54, 154, 275, 297, 132, 165, 110, 44, 10,  1,  0;
Central terms are Catalan numbers T(2*n,n) = A000108(n);
semi-diagonals form successive self-convolutions of the central terms:
  T(2*n+1,n) = [A000108^2](n),
  T(2*n+2,n) = [A000108^3](n).

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000108, A033184, A118340, A026010 (row sums shift left).

T:= proc(n, k) option remember;
      if k<0 or k>n then 0;
    elif k=0 then 1;
    elif k=n then 0;
    elif n>2*k then T(n, n-k) +T(n-1, k);
    else T(n, n-k-1) -T(n-1, k) -T(n-1, k+1);
      fi; end:
seq(seq(T(n, k), k=0..n), n=0..12); # G. C. Greubel, Mar 17 2021

T[n_, k_]:= T[n, k]= If[k<0 || k>n, 0, If[k==0, 1, If[k==n, 0, If[n>2*k, T[n, n-k] +T[n-1, k], T[n, n-k-1] -T[n-1, k] -T[n-1, k+1] ]]]];
Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* G. C. Greubel, Mar 17 2021 *)

T(n,k)=if(n2*k,T(n,n-k)+T(n-1,k),T(n,n-1-k)-T(n-1,k)-if(n-1>k,T(n-1,k+1)) ))))

@CachedFunction
def T(n, k):
    if (k<0 or k>n): return 0
    elif (k==0): return 1
    elif (k==n): return 0
    elif (n>2*k): return T(n, n-k) +T(n-1, k)
    else: return T(n, n-k-1) -T(n-1, k) -T(n-1, k+1)
flatten([[T(n, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Mar 17 2021

T(2*n+m, n) = [A000108^(m+1)](n), i.e., the m-th lower semi-diagonal forms the self-convolution (m+1)-power of A000108.

Sum_{k=0..n} T(n,k) = (1/2)*[n=0] + A026010(n-1) = (1/2)*[n=0] + (1/2)^((5 + (-1)^n)/2)*(6*n + 1 + 3*(-1)^n)*Catalan((2*n - 1 + (-1)^n)/4). - G. C. Greubel, Mar 17 2021

A096921 Triangle array of binomial coefficients.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 3, 3, 6, 1, 1, 3, 4, 6, 10, 1, 1, 4, 4, 10, 10, 20, 1, 1, 4, 5, 10, 15, 20, 35, 1, 1, 5, 5, 15, 15, 35, 35, 70, 1, 1, 5, 6, 15, 21, 35, 56, 70, 126, 1, 1, 6, 6, 21, 21, 56, 56, 126, 126, 252, 1, 1, 6, 7, 21, 28, 56, 84, 126, 210, 252, 462

Offset: 0

Paul Barry, Jul 15 2004

			Triangle begins:
      k=0  1  2  3  4  5
  n=0:  1;
  n=1:  1, 1;
  n=2:  1, 1, 2;
  n=3:  1, 1, 2, 3;
  n=4:  1, 1, 3, 3, 6;
  n=5:  1, 1, 3, 4, 6, 10;
  ...

Cf. A026010 (row sums), A016116 (diagonal sums), A001405 (main diagonal).

T[n_, k_]=Binomial[Floor[(n+k)/2], Floor[k/2]]; Table[T[n,k],{n,0,11},{k,0,n}] (* Stefano Spezia, Aug 23 2022 *)

T(n, k) = binomial((n+k)\2, k\2); \\ Michel Marcus, Oct 29 2022

T(n, k) = binomial(floor((n+k)/2), floor(k/2)).

A107249 A number triangle with repeated columns of binomial coefficients.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 2, 1, 1, 6, 3, 3, 2, 1, 1, 10, 6, 6, 3, 3, 1, 1, 20, 10, 10, 6, 4, 3, 1, 1, 35, 20, 20, 10, 10, 4, 4, 1, 1, 70, 35, 35, 20, 15, 10, 5, 4, 1, 1, 126, 70, 70, 35, 35, 15, 15, 5, 5, 1, 1, 252, 126, 126, 70, 56, 35, 21, 15, 6, 5, 1, 1, 462, 252, 252, 126, 126

Offset: 0

Paul Barry, May 14 2005

Reversal of A096921. Row sums are A026010. Diagonal sums are 107250.

			Triangle begins
1;
1,1;
1,1,1;
2,1,1,1;
3,2,2,1,1;
6,3,3,2,1,1;
10,6,6,3,3,1,1;

Number triangle T(n, k)=C(floor((2n-k-2)/2), floor((n-k)/2))

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A325699 Number of distinct even prime indices of n minus the number of distinct odd prime indices of n.

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A368175 Square array read by ascending antidiagonals: T(n,k) = Sum_{i=ceiling((k-n)/2)..floor((k+n-1)/2)} binomial(k,i), with n >= 1, k >= 0.

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A118344 Pendular Catalan triangle, read by rows, where row n is formed from row n-1 by the recurrence: if n > 2k, T(n,k) = T(n,n-k) + T(n-1,k), else T(n,k) = T(n,n-1-k) - T(n-1,k) - T(n-1,k+1), for n>=k>=0, with T(n,0)=1 and T(n,n)=0^n.

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A096921 Triangle array of binomial coefficients.

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A107249 A number triangle with repeated columns of binomial coefficients.

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