A137688 -id:A137688 - cp's OEIS frontend

A228196 A triangle formed like Pascal's triangle, but with n^2 on the left border and 2^n on the right border instead of 1.

0, 1, 2, 4, 3, 4, 9, 7, 7, 8, 16, 16, 14, 15, 16, 25, 32, 30, 29, 31, 32, 36, 57, 62, 59, 60, 63, 64, 49, 93, 119, 121, 119, 123, 127, 128, 64, 142, 212, 240, 240, 242, 250, 255, 256, 81, 206, 354, 452, 480, 482, 492, 505, 511, 512, 100, 287, 560, 806, 932, 962, 974, 997, 1016, 1023, 1024

Offset: 1

Boris Putievskiy, Aug 15 2013

The third row is (n^4 - n^2 + 24*n + 24)/12.

For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013

			The start of the sequence as a triangular array read by rows:
   0;
   1,  2;
   4,  3,  4;
   9,  7,  7,  8;
  16, 16, 14, 15, 16;
  25, 32, 30, 29, 31, 32;
  36, 57, 62, 59, 60, 63, 64;

Boris Putievskiy, Rows n = 1..140 of triangle, flattened
Rely Pellicer and David Alvo, Modified Pascal Triangle and Pascal Surfaces p.4
Index entries for triangles and arrays related to Pascal's triangle

Cf. We denote Pascal-like triangle with L(n) on the left border and R(n) on the right border by (L(n),R(n)). A007318 (1,1), A008949 (1,2^n), A029600 (2,3), A029618 (3,2), A029635 (1,2), A029653 (2,1), A037027 (Fibonacci(n),1), A051601 (n,n) n>=0, A051597 (n,n) n>0, A051666 (n^2,n^2), A071919 (1,0), A074829 (Fibonacci(n), Fibonacci(n)), A074909 (1,n), A093560 (3,1), A093561 (4,1), A093562 (5,1), A093563 (6,1), A093564 (7,1), A093565 (8,1), A093644 (9,1), A093645 (10,1), A095660 (1,3), A095666 (1,4), A096940 (1,5), A096956 (1,6), A106516 (3^n,1), A108561(1,(-1)^n), A132200 (4,4), A134636 (2n+1,2n+1), A137688 (2^n,2^n), A160760 (3^(n-1),1), A164844(1,10^n), A164847 (100^n,1), A164855 (101*100^n,1), A164866 (101^n,1), A172171 (1,9), A172185 (9,11), A172283 (-9,11), A177954 (int(n/2),1), A193820 (1,2^n), A214292 (n,-n), A227074 (4^n,4^n), A227075 (3^n,3^n), A227076 (5^n,5^n), A227550 (n!,n!), A228053 ((-1)^n,(-1)^n), A228074 (Fibonacci(n), n).

Cf. A000290 (row 1), A153056 (row 2), A000079 (column 1), A000225 (column 2), A132753 (column 3), A118885 (row sums of triangle array + 1), A228576 (generalized Pascal's triangle).

T:= function(n,k)
    if k=0 then return n^2;
    elif k=n then return 2^n;
    else return T(n-1,k-1) + T(n-1,k);
    fi;
  end;
Flat(List([0..12], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Nov 12 2019

T:= proc(n, k) option remember;
      if k=0 then n^2
    elif k=n then 2^k
    else T(n-1, k-1) + T(n-1, k)
      fi
    end:
seq(seq(T(n, k), k=0..n), n=0..10); # G. C. Greubel, Nov 12 2019

T[n_, k_]:= T[n, k] = If[k==0, n^2, If[k==n, 2^k, T[n-1, k-1] + T[n-1, k]]]; Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Nov 12 2019 *)
Flatten[Table[Sum[i^2 Binomial[n-1-i, n-k-i], {i,1,n-k}] + Sum[2^i Binomial[n-1-i, k-i], {i,1,k}], {n,0,10}, {k,0,n}]] (* Greg Dresden, Aug 06 2022 *)

T(n,k) = if(k==0, n^2, if(k==n, 2^k, T(n-1, k-1) + T(n-1, k) )); \\ G. C. Greubel, Nov 12 2019

def funcL(n):
   q = n**2
   return q
def funcR(n):
   q = 2**n
   return q
for n in range (1,9871):
   t=int((math.sqrt(8*n-7) - 1)/ 2)
   i=n-t*(t+1)/2-1
   j=(t*t+3*t+4)/2-n-1
   sum1=0
   sum2=0
   for m1 in range (1,i+1):
      sum1=sum1+funcR(m1)*binomial(i+j-m1-1,i-m1)
   for m2 in range (1,j+1):
      sum2=sum2+funcL(m2)*binomial(i+j-m2-1,j-m2)
   sum=sum1+sum2

@CachedFunction
def T(n, k):
    if (k==0): return n^2
    elif (k==n): return 2^n
    else: return T(n-1, k-1) + T(n-1, k)
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 12 2019

T(n,0) = n^2, n>0; T(0,k) = 2^k; T(n, k) = T(n-1, k-1) + T(n-1, k) for n,k > 0. [corrected by G. C. Greubel, Nov 12 2019]
Closed-form formula for general case. Let L(m) and R(m) be the left border and the right border of Pascal like triangle, respectively. We denote binomial(n,k) by C(n,k).
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} R(m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} L(m2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} R(m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} L(m2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2); n>0.
Some special cases. If L(m)={b,b,b...} b*A000012, then the second sum takes form b*C(n+k-1,j). If L(m) is {0,b,2b,...} b*A001477, then the second sum takes form b*C(n+k,n-1). Similarly for R(m) and the first sum.
For this sequence L(m)=m^2 and R(m)=2^m.
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} (2^m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} (m2^2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} (2^m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} (m2^2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2).
As a triangular array read by rows, T(n,k) = Sum_{i=1..n-k} i^2*C(n-1-i, n-k-i) + Sum_{i=1..k} 2^i*C(n-1-i, k-i); n,k >=0. - Greg Dresden, Aug 06 2022

Cross-references corrected and extended by Philippe Deléham, Dec 27 2013

A007664 Reve's puzzle: number of moves needed to solve the Towers of Hanoi puzzle with 4 pegs and n disks, according to the Frame-Stewart algorithm.

Original entry on oeis.org

0, 1, 3, 5, 9, 13, 17, 25, 33, 41, 49, 65, 81, 97, 113, 129, 161, 193, 225, 257, 289, 321, 385, 449, 513, 577, 641, 705, 769, 897, 1025, 1153, 1281, 1409, 1537, 1665, 1793, 2049, 2305, 2561, 2817, 3073, 3329, 3585, 3841, 4097, 4609, 5121, 5633

Offset: 0

N. J. A. Sloane, Mira Bernstein and Robert G. Wilson v

The Frame-Stewart algorithm minimizes the number of moves a(n) needed to first move k disks to an intermediate peg (requiring a(k) moves), then moving the remaining n-k disks to the destination peg without touching the k smallest disks (requiring 2^(n-k)-1 moves) and finally moving the k smaller disks to the destination.
This leads to the given recursive formula a(n) = min{...}. It follows that the sequence of first differences is A137688 = (1,2,2,4,4,4,...) = 2^A003056(n), which in turn gives the explicit formulas for a(n) as partial sums of A137688.
"Numerous others have rediscovered this algorithm over the years [several references omitted]; many of these failed to derive the correct value for the parameter i, most mistakenly thought that they had actually proved optimality and almost none contributed anything new to what was done by Frame and Stewart". [Stockmeyer]
Numbers of the form 2^k+1 appear for n = 2, 3, 4, 6, 8, 11, 15, 15+4 = 19, 19+5 = 24, 24+6 = 30, 30+7 = 37, 37+8 = 45, ... - Max Alekseyev, Feb 06 2008
The Frame-Stewart algorithm indeed gives the optimal solution, i.e., the minimal possible number of moves for the case of four pegs [Bousch, 2014]. - Andrey Zabolotskiy, Sep 18 2017

A. Brousseau, Tower of Hanoi with more pegs, J. Recreational Math., 8 (1975-1976), 169-176.
Paul Cull and E. F. Ecklund, On the Towers of Hanoi and generalized Towers of Hanoi problems. Proceedings of the thirteenth Southeastern conference on combinatorics, graph theory and computing (Boca Raton, Fla., 1982). Congr. Numer. 35 (1982), 229-238. MR0725883(85a:68059).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
D. Wood, Towers of Brahma and Hanoi revisited, J. Recreational Math., 14 (1981), 17-24.

Gheorghe Coserea, Table of n, a(n) for n = 0..10012 (first 1001 terms from M. F. Hasler)
S. Alejandre, Legend of Towers of Hanoi.
J.-P. Allouche, Note on the cyclic towers of Hanoi, Theoret. Comput. Sci., 123 (1994), 3-7.
T. Bousch, La quatrième tour de Hanoi, Bull. Belg. Math. Soc. Simon Stevin 21 (2014) 895-912.
A. Brousseau, Tower of Hanoi with more pegs, J. Recreational Math 8.3 (1975-6), 169-176. (Annotated scanned copy)
A. M. Hinz, An iterative algorithm for the Tower of Hanoi with four pegs, Computing, June 1989, Volume 42, Issue 2-3, pp. 133-140.
A. M. Hinz, S. Klavžar, U. Milutinović, and C. Petr, The Tower of Hanoi - Myths and Maths, Birkhäuser 2013.
B. Houston and H. Masum, Explorations in 4-peg Tower of Hanoi. [Paper]
B. Houston and H. Masum, Explorations in 4-peg Tower of Hanoi. [Web site]
S. Klavzar et al., Hanoi graphs and some classical numbers.
S. Klavzar and U. Milutinovic, Simple explicit formulas for the Frame-Stewart's numbers.
S. Klavzar, U. Milutinovic and C. Petr, On the Frame-Stewart algorithm for the multi-peg Tower of Hanoi problem, Discrete Appl. Math. 120, 1-3 (2002), 141 - 157.
Mathnet at U. Toronto, Generalizing the Towers of Hanoi Problem.
Richard E. Korf and Ariel Felner, Recent Progress in Heuristic Search: a Case Study of the Four-Peg Towers of Hanoi Problem. IJCAI 2007: 2324-2329.
B. M. Stewart, Advanced Problem 3918, Amer. Math. Monthly, 46 (1939), 363.
B. M. Stewart & J. S. Frame, Solution to Problem 3918, Amer. Math. Monthly, 48 (1941), 217-219.
P. Stockmeyer, Variations on the Four-Post Tower of Hanoi Puzzle, Congressus Numerantium 102 (1994), pp. 3-12. [Has extensive bibliography]
Eric Weisstein's World of Mathematics, Towers of Hanoi.
Janez Žerovnik, Self Similarities of the Tower of Hanoi Graphs and a proof of the Frame-Stewart Conjecture, arXiv:1601.04298 [math.CO], 2016.
Index entries for sequences related to Towers of Hanoi

Cf. A007665, A182058, A003056, A000225 (analog for 3 pegs), A137688 (first differences).

a007664 = sum . map (a000079 . a003056) . enumFromTo 0 . subtract 1
-- Reinhard Zumkeller, Feb 17 2013

A007664:=proc(n) option remember;min(seq(2*A007664(k)+2^(n-k)-1,k=0..n-1)) end; A007664(0):=0; # M. F. Hasler, Feb 06 2008
A007664 := n -> 1 + (n - 1 - A003056(n)*(A003056(n) - 1)/2)*2^A003056(n); A003056 := n -> round(sqrt(2*n+2))-1; # M. F. Hasler, Feb 06 2008

a[n_] := a[n] = Min[ Table[ 2*a[k] + 2^(n-k) - 1, {k, 0, n-1}]]; a[0] = 0; Table[a[n], {n, 0, 48}] (* Jean-François Alcover, Dec 06 2011, after M. F. Hasler *)
Join[{0},Accumulate[2^Flatten[Table[PadRight[{},n+1,n],{n,0,12}]]]] (* Harvey P. Dale, Jul 03 2021 *)

A007664(n) = (n - 1 - (n=A003056(n))*(n-1)/2)*2^n +1
A003056(n) = (sqrt(2*n+2)-.5)\1 \\ M. F. Hasler, Feb 06 2008

print_7664(n,s=0,t=1,c=1,d=1)=while(n-->=0,print1(s+=t,",");c--&next;c=d++;t<<=1)

A007664(n,c=1,d=1,t=1)=sum(i=c,n,i>c&(t<<=1)&c+=d++;t) \\ M. F. Hasler, Feb 06 2008

from math import isqrt
def A007664(n): return (1<<(r:=(k:=isqrt(m:=n+1<<1))+int(m>=k*(k+1)+1)-1))*(n-1-(r*(r-1)>>1))+1 # Chai Wah Wu, Oct 17 2022

a(n) = min{ 2 a(k) + 2^(n-k) - 1; k < n}, which is always odd. - M. F. Hasler, Feb 06 2008
a(n) = Sum_{i=0..n-1} 2^A003056(i). - Daniele Parisse, May 09 2003
a(n) = 1 + (n + A003056(n) - 1 - A003056(n)*(A003056(n) + 1)/2)*2^A003056(n). - Daniele Parisse, Feb 06 2001
a(n) = 1 + (n - 1 - A003056(n)*(A003056(n) - 1)/2)*2^A003056(n). - Daniele Parisse, Jul 07 2007

Edited, corrected and extended by M. F. Hasler, Feb 06 2008
Further edits by N. J. A. Sloane, Feb 08 2008
Upper bound updated with a reference by Max Alekseyev, Nov 23 2008

A140513 Repeat 2^n n times.

Original entry on oeis.org

2, 4, 4, 8, 8, 8, 16, 16, 16, 16, 32, 32, 32, 32, 32, 64, 64, 64, 64, 64, 64, 128, 128, 128, 128, 128, 128, 128, 256, 256, 256, 256, 256, 256, 256, 256, 512, 512, 512, 512, 512, 512, 512, 512, 512, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 1024

Offset: 0

Paul Curtz, Jul 01 2008

Reinhard Zumkeller, Rows n = 0..127 of triangle, flattened
Sajed Haque, Chapter 2.6.2 of Discriminators of Integer Sequences, 2017, See p. 34.

Cf. A000079, A018900, A059268, A111650, A137688.

a140513 n k = a140513_tabl !! (n-1) !! (k-1)
a140513_row n = a140513_tabl !! (n-1)
a140513_tabl = iterate (\xs@(x:_) -> map (* 2) (x:xs)) [2]
a140513_list = concat a140513_tabl
-- Reinhard Zumkeller, Nov 14 2015

t={}; Do[r={}; Do[If[k==0||k==n, m=2^n, m=t[[n, k]] + t[[n, k + 1]]]; r=AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t=Flatten[2 t] (* Vincenzo Librandi, Feb 17 2018 *)
Table[Table[2^n,n],{n,10}]//Flatten (* Harvey P. Dale, Dec 04 2018 *)

from math import isqrt
def A140513(n): return 1<<(m:=isqrt(k:=n+1<<1))+(k>m*(m+1)) # Chai Wah Wu, Nov 07 2024

a(n) = 2*A137688(n).
a(n) = A018900(n+1) - A059268(n). - Reinhard Zumkeller, Jun 24 2009
From Reinhard Zumkeller, Feb 28 2010: (Start)
Seen as a triangle read by rows: T(n,k)=2^n, 1 <= k <= n.
T(n,k) = A173786(n-1,k-1) + A173787(n-1,k-1), 1 <= k <= n. (End)
Sum_{n>=0} 1/a(n) = 2. - Amiram Eldar, Aug 16 2022

A227075 A triangle formed like Pascal's triangle, but with 3^n on the borders instead of 1.

Original entry on oeis.org

1, 3, 3, 9, 6, 9, 27, 15, 15, 27, 81, 42, 30, 42, 81, 243, 123, 72, 72, 123, 243, 729, 366, 195, 144, 195, 366, 729, 2187, 1095, 561, 339, 339, 561, 1095, 2187, 6561, 3282, 1656, 900, 678, 900, 1656, 3282, 6561, 19683, 9843, 4938, 2556, 1578, 1578, 2556, 4938

Offset: 0

T. D. Noe, Aug 01 2013

All rows except the zeroth are divisible by 3. Is there a closed-form formula for these numbers, like for binomial coefficients?

Let b=3 and T(n,k) = A(n-k,k) be the associated reading of the symmetric array A by antidiagonals, then A(n,k) = sum_{r=1..n} b^r*A178300(n-r,k) + sum_{c=1..k} b^c*A178300(k-c,n). Similarly with b=4 and b=5 for A227074 and A227076. - R. J. Mathar, Aug 10 2013

			Triangle:
1,
3, 3,
9, 6, 9,
27, 15, 15, 27,
81, 42, 30, 42, 81,
243, 123, 72, 72, 123, 243,
729, 366, 195, 144, 195, 366, 729,
2187, 1095, 561, 339, 339, 561, 1095, 2187,
6561, 3282, 1656, 900, 678, 900, 1656, 3282, 6561

T. D. Noe, Rows n = 0..50 of triangle, flattened

Cf. A007318 (Pascal's triangle), A228053 ((-1)^n on the borders).
Cf. A051601 (n on the borders), A137688 (2^n on borders).
Cf. A166060 (row sums: 4*3^n - 3*2^n), A227074 (4^n edges), A227076 (5^n edges).

t = {}; Do[r = {}; Do[If[k == 0 || k == n, m = 3^n, m = t[[n, k]] + t[[n, k + 1]]]; r = AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t = Flatten[t]

A227550 A triangle formed like Pascal's triangle, but with factorial(n) on the borders instead of 1.

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 6, 4, 4, 6, 24, 10, 8, 10, 24, 120, 34, 18, 18, 34, 120, 720, 154, 52, 36, 52, 154, 720, 5040, 874, 206, 88, 88, 206, 874, 5040, 40320, 5914, 1080, 294, 176, 294, 1080, 5914, 40320, 362880, 46234, 6994, 1374, 470, 470, 1374, 6994, 46234, 362880, 3628800

Offset: 0

Vincenzo Librandi, Aug 04 2013

A003422 gives the second column (after 0).

			Triangle begins:
       1;
       1,     1;
       2,     2,    2;
       6,     4,    4,    6;
      24,    10,    8,   10,  24;
     120,    34,   18,   18,  34, 120;
     720,   154,   52,   36,  52, 154,  720;
    5040,   874,  206,   88,  88, 206,  874, 5040;
   40320,  5914, 1080,  294, 176, 294, 1080, 5914, 40320;
  362880, 46234, 6994, 1374, 470, 470, 1374, 6994, 46234, 362880;

Vincenzo Librandi, Rows n = 0..70, flattened

Cf. similar triangles with t on the borders: A007318 (t = 1), A028326 (t = 2), A051599 (t = prime(n)), A051601 (t = n), A051666 (t = n^2), A108617 (t = fibonacci(n)), A134636 (t = 2n+1), A137688 (t = 2^n), A227075 (t = 3^n).
Cf. A003422.
Cf. A227791 (central terms), A001563, A074911.

a227550 n k = a227550_tabl !! n !! k
a227550_row n = a227550_tabl !! n
a227550_tabl = map fst $ iterate
   (\(vs, w:ws) -> (zipWith (+) ([w] ++ vs) (vs ++ [w]), ws))
   ([1], a001563_list)
-- Reinhard Zumkeller, Aug 05 2013

function T(n,k)
  if k eq 0 or k eq n then return Factorial(n);
  else return T(n-1,k-1) + T(n-1,k);
  end if; return T;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, May 02 2021

t = {}; Do[r = {}; Do[If[k == 0||k == n, m = n!, m = t[[n, k]] + t[[n, k + 1]]]; r = AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t = Flatten[t]

def T(n,k): return factorial(n) if (k==0 or k==n) else T(n-1, k-1) + T(n-1, k)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 02 2021

From G. C. Greubel, May 02 2021: (Start)
T(n, k) = T(n-1, k-1) + T(n-1, k) with T(n, 0) = T(n, n) = n!.
Sum_{k=0..n} T(n, k) = 2^n * (1 +Sum_{j=1..n-1} j*j!/2^j) = A140710(n). (End)

A228053 A triangle formed like Pascal's triangle, but with (-1)^(n+1) on the borders instead of 1.

Original entry on oeis.org

-1, 1, 1, -1, 2, -1, 1, 1, 1, 1, -1, 2, 2, 2, -1, 1, 1, 4, 4, 1, 1, -1, 2, 5, 8, 5, 2, -1, 1, 1, 7, 13, 13, 7, 1, 1, -1, 2, 8, 20, 26, 20, 8, 2, -1, 1, 1, 10, 28, 46, 46, 28, 10, 1, 1, -1, 2, 11, 38, 74, 92, 74, 38, 11, 2, -1, 1, 1, 13, 49, 112, 166, 166, 112

Offset: 0

T. D. Noe, Aug 07 2013

This sequence is almost the same as A026637.

T(n,k) = A026637(n-2,k-1) for n > 3, 1 < k < n-1. - Reinhard Zumkeller, Aug 08 2013

			Triangle begins:
  -1,
  1, 1,
  -1, 2, -1,
  1, 1, 1, 1,
  -1, 2, 2, 2, -1,
  1, 1, 4, 4, 1, 1,
  -1, 2, 5, 8, 5, 2, -1,
  1, 1, 7, 13, 13, 7, 1, 1,
  -1, 2, 8, 20, 26, 20, 8, 2, -1,
  1, 1, 10, 28, 46, 46, 28, 10, 1, 1,
  -1, 2, 11, 38, 74, 92, 74, 38, 11, 2, -1

T. D. Noe, Rows n = 0..50 of triangle, flattened

Cf. A007318 (Pascal's triangle), A026637 (many terms in common).
Cf. A051601 (n on the borders), A137688 (2^n on borders).
Cf. A097073 (row sums).
Cf. A227074 (4^n edges), A227075 (3^n edges), A227076 (5^n edges).

a228053 n k = a228053_tabl !! n !! k
a228053_row n = a228053_tabl !! n
a228053_tabl = iterate (\row@(i:_) -> zipWith (+)
   ([- i] ++ tail row ++ [0]) ([0] ++ init row ++ [- i])) [- 1]
  -- Reinhard Zumkeller, Aug 08 2013

t = {}; Do[r = {}; Do[If[k == 0 || k == n, m = (-1)^(n+1), m = t[[n, k]] + t[[n, k + 1]]]; r = AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t = Flatten[t]

A227074 A triangle formed like Pascal's triangle, but with 4^n on the borders instead of 1.

Original entry on oeis.org

1, 4, 4, 16, 8, 16, 64, 24, 24, 64, 256, 88, 48, 88, 256, 1024, 344, 136, 136, 344, 1024, 4096, 1368, 480, 272, 480, 1368, 4096, 16384, 5464, 1848, 752, 752, 1848, 5464, 16384, 65536, 21848, 7312, 2600, 1504, 2600, 7312, 21848, 65536, 262144, 87384, 29160

Offset: 0

T. D. Noe, Aug 06 2013

All rows except the zeroth are divisible by 4. Is there a closed-form formula for these numbers, like for binomial coefficients?

			Triangle begins:
  1,
  4, 4,
  16, 8, 16,
  64, 24, 24, 64,
  256, 88, 48, 88, 256,
  1024, 344, 136, 136, 344, 1024,
  4096, 1368, 480, 272, 480, 1368, 4096,
  16384, 5464, 1848, 752, 752, 1848, 5464, 16384,
  65536, 21848, 7312, 2600, 1504, 2600, 7312, 21848, 65536

T. D. Noe, Rows n = 0..50 of triangle, flattened

Cf. A007318 (Pascal's triangle), A228053 ((-1)^n on the borders).
Cf. A051601 (n on the borders), A137688 (2^n on borders).
Cf. A165665 (row sums: 3*4^n - 2*2^n), A227075 (3^n edges), A227076 (5^n edges).

t = {}; Do[r = {}; Do[If[k == 0 || k == n, m = 4^n, m = t[[n, k]] + t[[n, k + 1]]]; r = AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t = Flatten[t]

A227076 A triangle formed like Pascal's triangle, but with 5^n on the borders instead of 1.

Original entry on oeis.org

1, 5, 5, 25, 10, 25, 125, 35, 35, 125, 625, 160, 70, 160, 625, 3125, 785, 230, 230, 785, 3125, 15625, 3910, 1015, 460, 1015, 3910, 15625, 78125, 19535, 4925, 1475, 1475, 4925, 19535, 78125, 390625, 97660, 24460, 6400, 2950, 6400, 24460, 97660, 390625

Offset: 0

T. D. Noe, Aug 06 2013

All rows except the zeroth are divisible by 5. Is there a closed-form formula for these numbers, as there is for binomial coefficients?

			Triangle begins as:
       1;
       5,     5;
      25,    10,    25;
     125,    35,    35,  125;
     625,   160,    70,  160,  625;
    3125,   785,   230,  230,  785, 3125;
   15625,  3910,  1015,  460, 1015, 3910, 15625;
   78125, 19535,  4925, 1475, 1475, 4925, 19535, 78125;
  390625, 97660, 24460, 6400, 2950, 6400, 24460, 97660, 390625;

T. D. Noe, Rows n = 0..50 of triangle, flattened

Cf. A007318 (Pascal's triangle), A228053 ((-1)^n on the borders).
Cf. A051601 (n on the borders), A137688 (2^n on borders).
Cf. A083585 (row sums: (8*5^n - 5*2^n)/3), A227074 (4^n edges), A227075 (3^n edges).
Cf. A000351.

function T(n,k) // T = A227076
  if k eq 0 or k eq n then return 5^n;
  else return T(n-1,k) + T(n-1,k-1);
  end if;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 10 2025

A227076 := proc(n,k)
    if k = 0 or k = n then
        5^n ;
    elif k < 0 or k > n then
        0;
    else
        procname(n-1,k)+procname(n-1,k-1) ;
    end if;
end proc: # R. J. Mathar, Aug 09 2013

t = {}; Do[r = {}; Do[If[k == 0 || k == n, m = 5^n, m = t[[n, k]] + t[[n, k + 1]]]; r = AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t = Flatten[t]

from sage.all import *
@CachedFunction
def T(n,k): # T = A227076
    if k==0 or k==n: return pow(5,n)
    else: return T(n-1,k) + T(n-1,k-1)
print(flatten([[T(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Jan 10 2025

From R. J. Mathar, Aug 09 2013: (Start)
T(n,0) = 5^n.
T(n,1) = 5*A047850(n-1).
T(n,2) = 5*(5^n/80 + 3*n/4 + 51/16).
T(n,3) = 5*(5^n/320 + 45*n/16 + 3*n^2/8 + 819/64). (End)
Sum_{k=0..n} (-1)^k*T(n, k) = 20*(1+(-1)^n)*A009969(floor((n-1)/2)) - (3/5)*[n = 0]. - G. C. Greubel, Jan 10 2025

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