A028552 -id:A028552 - cp's OEIS frontend

A132764 a(n) = n*(n+22).

0, 23, 48, 75, 104, 135, 168, 203, 240, 279, 320, 363, 408, 455, 504, 555, 608, 663, 720, 779, 840, 903, 968, 1035, 1104, 1175, 1248, 1323, 1400, 1479, 1560, 1643, 1728, 1815, 1904, 1995, 2088, 2183, 2280, 2379, 2480, 2583, 2688, 2795, 2904, 3015, 3128, 3243, 3360

Offset: 0

Omar E. Pol, Aug 28 2007

			a(1)=2*1+0+21=23; a(2)=2*2+23+21=48; a(3)=2*3+48+21=75. - _Vincenzo Librandi_, Aug 03 2010

G. C. Greubel, Table of n, a(n) for n = 0..5000
Felix P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, Preprint on ResearchGate, March 2014.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001008, A002378, A005563, A028347, A028552, A028557, A028560, A028563, A028566, A028569, A098603, A098847, A098848, A098849, A098850, A102928, A120071, A132759, A132760, A132761, A132762, A132763, A132765, A132766, A132767.

Table[n(n+22),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,23,48},50] (* Harvey P. Dale, May 02 2012 *)

a(n)=n*(n+22) \\ Charles R Greathouse IV, Oct 07 2015

[n*(n+22) for n in (0..50)] # G. C. Greubel, Mar 14 2022

a(n) = n*(n + 22).
a(n) = 2*n + a(n-1) + 21 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
a(0)=0, a(1)=23, a(2)=48, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, May 02 2012
From Amiram Eldar, Jan 16 2021: (Start)
Sum_{n>=1} 1/a(n) = H(22)/22 = A001008(22)/A102928(22) = 19093197/113809696, where H(k) is the k-th harmonic number.
Sum_{n>=1} (-1)^(n+1)/a(n) = 156188887/5121436320. (End)
From G. C. Greubel, Mar 14 2022: (Start)
G.f.: x*(23 - 21*x)/(1-x)^3.
E.g.f.: x*(23 + x)*exp(x). (End)

A132763 a(n) = n*(n+21).

Original entry on oeis.org

0, 22, 46, 72, 100, 130, 162, 196, 232, 270, 310, 352, 396, 442, 490, 540, 592, 646, 702, 760, 820, 882, 946, 1012, 1080, 1150, 1222, 1296, 1372, 1450, 1530, 1612, 1696, 1782, 1870, 1960, 2052, 2146, 2242, 2340, 2440, 2542, 2646, 2752, 2860, 2970, 3082, 3196, 3312

Offset: 0

Omar E. Pol, Aug 28 2007

G. C. Greubel, Table of n, a(n) for n = 0..5000
Felix P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, Preprint on ResearchGate, March 2014.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001008, A002378, A005563, A028347, A028552, A028557, A028560, A028563, A028566, A028569, A098603, A098847, A098848, A098849, A098850, A102928, A120071, A132759, A132760, A132761, A132762, A132764, A132765, A132766, A132767.

Table[n(n+21),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,22,46},50] (* Harvey P. Dale, May 25 2014 *)

a(n)=n*(n+21) \\ Charles R Greathouse IV, Oct 07 2015

[n*(n+21) for n in (0..50)] # G. C. Greubel, Mar 14 2022

a(n) = n*(n + 21).
a(n) = 2*n + a(n-1) + 20 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
a(0)=0, a(1)=22, a(2)=46, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, May 25 2014
From Amiram Eldar, Jan 16 2021: (Start)
Sum_{n>=1} 1/a(n) = H(21)/21 = A001008(21)/A102928(21) = 18858053/108636528, where H(k) is the k-th harmonic number.
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*log(2)/21 - 166770367/4888643760. (End)
From Stefano Spezia, Jan 30 2021: (Start)
O.g.f.: 2*x*(11 - 10*x)/(1 - x)^3.
E.g.f.: x*(22 + x)*exp(x). (End)

A132766 a(n) = n*(n+24).

Original entry on oeis.org

0, 25, 52, 81, 112, 145, 180, 217, 256, 297, 340, 385, 432, 481, 532, 585, 640, 697, 756, 817, 880, 945, 1012, 1081, 1152, 1225, 1300, 1377, 1456, 1537, 1620, 1705, 1792, 1881, 1972, 2065, 2160, 2257, 2356, 2457, 2560, 2665, 2772, 2881, 2992, 3105, 3220, 3337

Offset: 0

Omar E. Pol, Aug 28 2007

G. C. Greubel, Table of n, a(n) for n = 0..5000
Felix P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, Preprint on ResearchGate, March 2014.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001008, A002378, A005563, A028347, A028552, A028557, A028560, A028563, A028566, A028569, A098603, A098849, A098850, A098847, A098848, A102928, A120071, A132759, A132760, A132761, A132762, A132763, A132764, A132765.

Table[n (n + 24), {n, 0, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 25, 52}, 50] (* Harvey P. Dale, Feb 11 2016 *)

a(n)=n*(n+24) \\ Charles R Greathouse IV, Jun 17 2017

[n*(n+24) for n in (0..50)] # G. C. Greubel, Mar 14 2022

a(n) = n*(n + 24).
a(n) = 2*n + a(n-1) + 23 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
a(0)=0, a(1)=25, a(2)=52; for n>2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Feb 11 2016
From Amiram Eldar, Jan 16 2021: (Start)
Sum_{n>=1} 1/a(n) = H(24)/24 = A001008(24)/A102928(24) = 1347822955/8566766208, where H(k) is the k-th harmonic number.
Sum_{n>=1} (-1)^(n+1)/a(n) = 3602044091/128501493120. (End)
From G. C. Greubel, Mar 14 2022: (Start)
G.f.: 2*x*(13 - 12*x)/(1-x)^3.
E.g.f.: x*(26 + x)*exp(x). (End)

A132767 a(n) = n*(n + 25).

Original entry on oeis.org

0, 26, 54, 84, 116, 150, 186, 224, 264, 306, 350, 396, 444, 494, 546, 600, 656, 714, 774, 836, 900, 966, 1034, 1104, 1176, 1250, 1326, 1404, 1484, 1566, 1650, 1736, 1824, 1914, 2006, 2100, 2196, 2294, 2394, 2496, 2600, 2706, 2814, 2924, 3036, 3150, 3266, 3384

Offset: 0

Omar E. Pol, Aug 28 2007

a(n) is the Zagreb 1 index of the Mycielskian of the cycle graph C[n]. See p. 205 of the D. B. West reference. - Emeric Deutsch, Nov 04 2016

Douglas B. West, Introduction to Graph Theory, 2nd ed., Prentice-Hall, NJ, 2001.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Felix P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, Preprint on ResearchGate, March 2014.
Wikipedia, Mycielskian.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001008, A002378, A005563, A028347, A028552, A028557, A028560, A028563, A028566, A028569, A098603, A098847, A098848, A098849, A098850, A102928, A120071, A132759, A132760, A132761, A132762, A132763, A132764, A132765, A132766.

Table[n (n + 25), {n, 0, 50}] (* Bruno Berselli, Aug 22 2018 *)
LinearRecurrence[{3,-3,1},{0,26,54},60] (* Harvey P. Dale, Feb 20 2023 *)

a(n)=n*(n+25) \\ Charles R Greathouse IV, Jun 17 2017

[n*(n+25) for n in (0..50)] # G. C. Greubel, Mar 13 2022

a(n) = 2*n + a(n-1) + 24 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
a(n) = n^2 + 25*n. - Omar E. Pol, Nov 04 2016
From Chai Wah Wu, Dec 17 2016: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2.
G.f.: 2*x*(13 - 12*x)/(1-x)^3. (End)
From Amiram Eldar, Jan 16 2021: (Start)
Sum_{n>=1} 1/a(n) = H(25)/25 = A001008(25)/A102928(25) = 34052522467/223092870000, where H(k) is the k-th harmonic number.
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*log(2)/25 - 19081066231/669278610000. (End)
E.g.f.: x*(26 + x)*exp(x). - G. C. Greubel, Mar 13 2022

A064801 Take 1, skip 2, take 2, skip 3, take 3, etc.

Original entry on oeis.org

1, 4, 5, 9, 10, 11, 16, 17, 18, 19, 25, 26, 27, 28, 29, 36, 37, 38, 39, 40, 41, 49, 50, 51, 52, 53, 54, 55, 64, 65, 66, 67, 68, 69, 70, 71, 81, 82, 83, 84, 85, 86, 87, 88, 89, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 121, 122, 123, 124, 125, 126, 127, 128

Offset: 1

Robert G. Wilson v, Oct 21 2001

A253607(a(n)) < 0. - Reinhard Zumkeller, Jan 05 2015
Integers m such that A000196(m) = A079643(m). - Firas Melaih, Dec 10 2020
Also possible values of floor(x*floor(x)) for real x >= 1. - Jianing Song, Feb 16 2021

Harry J. Smith, Table of n, a(n) for n = 1..1000

Cf. A007606, A004202, A048859.
Cf. A061885 (complement), A253607.
Cf. A136272.

a064801 n = a064801_list !! (n-1)
a064801_list = f 1 [1..] where
   f k xs = us ++ f (k + 1) (drop (k + 1) vs)
            where (us, vs) = splitAt k xs
-- Reinhard Zumkeller, May 16 2014

seq(`if`(floor(sqrt(k)) * (floor(sqrt(k)) + 1) > k, k, NULL), k = 0..2034); # a(1)..a(1000), Rainer Rosenthal, Jul 19 2024

a = Table[n, {n, 0, 200} ]; b = {}; Do[a = Drop[a, {1, n} ]; b = Append[b, Take[a, {1, n} ]]; a = Drop[a, {1, n} ], {n, 1, 14} ]; Flatten[b]
Flatten[Table[Range[n^2,n^2+n-1],{n,12}]] (* Harvey P. Dale, Dec 18 2015 *)

{ n=0; for (m=1, 10^9, s=m^2; a=0; for (k=0, m - 1, a=s+k; write("b064801.txt", n++, " ", a); if (n==1000, return)) ) } \\ Harry J. Smith, Sep 26 2009

from math import isqrt  # after Rainer Rosenthal
def isA(k: int): return k < ((s:=isqrt(k)) * (s + 1))
print([k for k in range(129) if isA(k)]) # Peter Luschny, Jul 19 2024

a(n) = A004202(n) - 1.
Can be interpreted as a table read by rows: T(n,k) = n^2 + k, 0 <= k < n. T(n,k) = 0 iff k > A000196(n); T(n,0) = A000290(n); T(n,1) = A002522(n) for n > 1; T(n,2) = A010000(n) = A059100(n) for n > 2; T(n, n-3) = A014209(n-1) for n > 2; T(n, n-2) = A028552(n) for n > 1; T(n, n-1) = A028387(n-1); T(2*n+1, n) = A001107(n+1). - Reinhard Zumkeller, Nov 18 2003
Numbers k such that floor(sqrt(k)) * (floor(sqrt(k)) + 1) > k. - Rainer Rosenthal, Jul 19 2024

A100345 Triangle read by rows: T(n,k) = n*(n+k), 0 <= k <= n.

Original entry on oeis.org

0, 1, 2, 4, 6, 8, 9, 12, 15, 18, 16, 20, 24, 28, 32, 25, 30, 35, 40, 45, 50, 36, 42, 48, 54, 60, 66, 72, 49, 56, 63, 70, 77, 84, 91, 98, 64, 72, 80, 88, 96, 104, 112, 120, 128, 81, 90, 99, 108, 117, 126, 135, 144, 153, 162, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190

Offset: 0

Reinhard Zumkeller, Nov 18 2004

Distinct members (except 0) are in A071562. Numbers occurring at least twice are in A175040. - Franklin T. Adams-Watters, Apr 04 2010

			Triangle begins:
   0
   1   2
   4   6   8
   9  12  15  18
  16  20  24  28  32
  25  30  35  40  45  50
  36  42  48  54  60  66  72
  49  56  63  70  77  84  91  98
  64  72  80  88  96 104 112 120 128

Cf. A000290, A000384, A001105, A002378, A005563, A014107, A028347, A028552, A028557, A028560, A028563, A028566, A028569, A046092, A054000, A071355, A071562, A085789, A098603, A175040.

Table[n(n+k),{n,0,10},{k,0,n}]//Flatten (* Harvey P. Dale, Aug 16 2018 *)

row(n) = vector(n+1, k, n*(n+k-1)); \\ Amiram Eldar, May 09 2025

T(n,0) = A000290(n).
T(n,1) = A002378(n) for n > 0.
T(n,2) = A005563(n) for n > 1.
T(n,3) = A028552(n) for n > 2.
T(n,4) = A028347(n+2) for n > 3.
T(n,5) = A028557(n) for n > 4.
T(n,6) = A028560(n) for n > 5.
T(n,7) = A028563(n) for n > 6.
T(n,8) = A028566(n) for n > 7.
T(n,9) = A028569(n) for n > 8.
T(n,10) = A098603(n) for n > 9.
T(n,n-5) = A071355(n-4) for n > 4.
T(n,n-4) = A054000(n-1) for n > 3.
T(n,n-3) = A014107(n) for n > 2.
T(n,n-2) = A046092(n-1) for n > 1.
T(n,n-1) = A000384(n) for n > 0.
T(n,n) = A001105(n).
Row sums give A085789 for n > 0.
G.f.: x*(1 + 2*y + 6*x^3*y^2 - 3*x^2*y*(1 + 2*y) + x*(1 - 3*y + 2*y^2))/((1 - x)^3*(1 - x*y)^3). - Stefano Spezia, Jul 03 2025

A105937 Infinite square array read by antidiagonals: T(m, 0) = 1, T(m, 1) = m; T(m, k) = (m - k + 1) T(m+1, k-1) - (k-1) (m+1) T(m+2, k-2).

Original entry on oeis.org

1, 1, 0, 1, 1, -2, 1, 2, -2, 0, 1, 3, 0, -12, 36, 1, 4, 4, -24, 24, 0, 1, 5, 10, -30, -60, 420, -1800, 1, 6, 18, -24, -216, 720, -720, 0, 1, 7, 28, 0, -420, 420, 5040, -30240, 176400, 1, 8, 40, 48, -624, -960, 14400, -40320, 40320, 0, 1, 9, 54, 126, -756, -3780, 22680, 22680, -589680, 3764880, -28576800

Offset: 0

Vincent v.d. Noort, Mar 24 2007

			Array begins
   1  1  1   1   1   1   1   1   1   1 ... (A000012)
   0  1  2   3   4   5   6   7   8   9 ... (A001477)
  -2 -2  0   4  10  18  28  40  54  70 ... (A028552)
   0 12 24  30  24   0  48 126 240 396 ... (A126935)
  36 24 60 216 420 624 756 720 396 360 ... (A126958)
...

V. van der Noort and N. J. A. Sloane, Paper in preparation, 2007.

G. C. Greubel, Antidiagonals n = 0..100, flattened

Rows give A000027, A028552, A126935, A126958.
Columns give A126934, A126962, A127067, A127068, A127070.
A127080 gives another version of the array.

function T(n,k)
  if k eq 0 then return 1;
  elif k eq 1 then return n;
  else return (n-k+1)*T(n+1, k-1) - (k-1)*(n+1)*T(n+2, k-2);
  end if; return T; end function;
[T(n-k,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 28 2020

T:= proc(n, k) option remember;
      if k=0 then 1
    elif k=1 then n
    else (n-k+1)*T(n+1, k-1) - (k-1)*(n+1)*T(n+2, k-2)
      fi; end:
seq(seq(T(n-k, k), k=0..n), n=0..12); # G. C. Greubel, Jan 28 2020

T[n_, k_]:= T[n, k]= If[k==0, 1, If[k==1, n, (n-k+1)*T[n+1, k-1] - (k-1)*(n+1)* T[n+2, k-2]]]; Table[T[n-k,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 28 2020 *)

T(n,k) = if(k==0, 1, if(k==1, n, (n-k+1)*T(n+1, k-1) - (k-1)*(n+1)*T(n+2, k-2) )); \\ G. C. Greubel, Jan 28 2020

@CachedFunction
def T(n, k):
    if (k==0): return 1
    elif (k==1): return n
    else: return (n-k+1)*T(n+1, k-1) - (k-1)*(n+1)*T(n+2, k-2)
[[T(n-k, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jan 28 2020

See A127080 for e.g.f.

More terms added by G. C. Greubel, Jan 28 2020

A285595 Sum T(n,k) of the k-th entries in all blocks of all set partitions of [n]; triangle T(n,k), n>=1, 1<=k<=n, read by rows.

Original entry on oeis.org

1, 4, 2, 17, 10, 3, 76, 52, 18, 4, 362, 274, 111, 28, 5, 1842, 1500, 675, 200, 40, 6, 9991, 8614, 4185, 1380, 325, 54, 7, 57568, 51992, 26832, 9568, 2510, 492, 70, 8, 351125, 329650, 178755, 67820, 19255, 4206, 707, 88, 9, 2259302, 2192434, 1239351, 494828, 149605, 35382, 6629, 976, 108, 10

Offset: 1

Alois P. Heinz, Apr 22 2017

T(n,k) is also k times the number of blocks of size >k in all set partitions of [n+1]. T(3,2) = 10 = 2 * 5 because there are 5 blocks of size >2 in all set partitions of [4], namely in 1234, 123|4, 124|3, 134|2, 1|234.

			T(3,2) = 10 because the sum of the second entries in all blocks of all set partitions of [3] (123, 12|3, 13|2, 1|23, 1|2|3) is 2+2+3+3+0  = 10.
Triangle T(n,k) begins:
      1;
      4,     2;
     17,    10,     3;
     76,    52,    18,    4;
    362,   274,   111,   28,    5;
   1842,  1500,   675,  200,   40,   6;
   9991,  8614,  4185, 1380,  325,  54,  7;
  57568, 51992, 26832, 9568, 2510, 492, 70, 8;
  ...

Alois P. Heinz, Rows n = 1..141, flattened
Wikipedia, Partition of a set

Column k=1 gives A124325(n+1).
Row sums give A000110(n) * A000217(n) = A105488(n+3).
Main diagonal and first lower diagonal give: A000027, A028552.
Cf. A007318, A175757, A278677, A283424, A285362, A285793, A286897.

T:= proc(h) option remember; local b; b:=
      proc(n, l) option remember; `if`(n=0, [1, 0],
        (p-> p+[0, (h-n+1)*p[1]*x^1])(b(n-1, [l[], 1]))+
         add((p-> p+[0, (h-n+1)*p[1]*x^(l[j]+1)])(b(n-1,
         sort(subsop(j=l[j]+1, l), `>`))), j=1..nops(l)))
      end: (p-> seq(coeff(p, x, i), i=1..n))(b(h, [])[2])
    end:
seq(T(n), n=1..12);
# second Maple program:
b:= proc(n) option remember; `if`(n=0, [1, 0],
      add((p-> p+[0, p[1]*add(x^k, k=1..j-1)])(
         b(n-j)*binomial(n-1, j-1)), j=1..n))
    end:
T:= n-> (p-> seq(coeff(p, x, i)*i, i=1..n))(b(n+1)[2]):
seq(T(n), n=1..12);

b[n_] := b[n] = If[n == 0, {1, 0}, Sum[# + {0, #[[1]]*Sum[x^k, {k, 1, j-1} ]}&[b[n - j]*Binomial[n - 1, j - 1]], {j, 1, n}]];
T[n_] := Table[Coefficient[#, x, i]*i, {i, 1, n}] &[b[n + 1][[2]]];
Table[T[n], {n, 1, 12}] // Flatten (* Jean-François Alcover, May 23 2018, translated from 2nd Maple program *)

T(n,k) = k * Sum_{j=k+1..n+1} binomial(n+1,j)*A000110(n+1-j).
T(n,k) = k * Sum_{j=k+1..n+1} A175757(n+1,j).
Sum_{k=1..n} T(n,k)/k = A278677(n-1).

A132768 a(n) = n*(n + 26).

Original entry on oeis.org

0, 27, 56, 87, 120, 155, 192, 231, 272, 315, 360, 407, 456, 507, 560, 615, 672, 731, 792, 855, 920, 987, 1056, 1127, 1200, 1275, 1352, 1431, 1512, 1595, 1680, 1767, 1856, 1947, 2040, 2135, 2232, 2331, 2432, 2535, 2640, 2747, 2856, 2967, 3080, 3195, 3312, 3431

Offset: 0

Omar E. Pol, Aug 28 2007

G. C. Greubel, Table of n, a(n) for n = 0..5000
Felix P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, Preprint on ResearchGate, March 2014.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001008, A002378, A005563, A028347, A028552, A028557, A028560, A028563, A028566, A028569.
Cf. A098849, A098850, A098603, A098847, A098848, A102928, A120071, A132759, A132760, A132761.
Cf. A132762, A132763, A132764, A132765, A132766, A132767.

Table[n(n+26),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,27,56},50] (* Harvey P. Dale, Dec 15 2018 *)

a(n)=n*(n+26) \\ Charles R Greathouse IV, Jun 17 2017

[n*(n+26) for n in (0..50)] # G. C. Greubel, Mar 13 2022

a(n) = n*(n + 26).
a(n) = 2*n + a(n-1) + 25, with a(0)=0. - Vincenzo Librandi, Aug 03 2010
From Amiram Eldar, Jan 16 2021: (Start)
Sum_{n>=1} 1/a(n) = H(26)/26 = A001008(26)/A102928(26) = 34395742267/232016584800, where H(k) is the k-th harmonic number.
Sum_{n>=1} (-1)^(n+1)/a(n) = 18051406831/696049754400. (End)
From G. C. Greubel, Mar 13 2022: (Start)
G.f.: x*(27 - 25*x)/(1-x)^3.
E.g.f.: x*(27 + x)*exp(x). (End)

A258219 A(n,k) is the sum over all Dyck paths of semilength n of products over all peaks p of (x_p+k*y_p)/y_p, where x_p and y_p are the coordinates of peak p; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 4, 1, 3, 10, 25, 1, 4, 18, 74, 208, 1, 5, 28, 153, 706, 2146, 1, 6, 40, 268, 1638, 8162, 26368, 1, 7, 54, 425, 3172, 20898, 110410, 375733, 1, 8, 70, 630, 5500, 44164, 307908, 1708394, 6092032, 1, 9, 88, 889, 8838, 82850, 702844, 5134293, 29752066, 110769550

Offset: 0

Alois P. Heinz, May 23 2015

A Dyck path of semilength n is a (x,y)-lattice path from (0,0) to (2n,0) that does not go below the x-axis and consists of steps U=(1,1) and D=(1,-1). A peak of a Dyck path is any lattice point visited between two consecutive steps UD.

Conjecture: the g.f. G(k,x) for the k-th column satisfies the Riccati differential equation 2*x^2*d/dx(G(k,x)) + 1 + (k*x - 1)*G(k,x) + x*G^2(k,x) = 0 and hence, by Stokes 1982, has the continued fraction representation G(k,x) = 1/(1 - (k+1)*x/(1 - 3*x/(1 - (k+3)*x/(1 - 5*x/(1 - (k+5)*x/(1 - 7*x/(1 - ...))))))) of Stieltjes type. - Peter Bala, Jul 28 2022

			Square array A(n,k) begins:
     1,    1,     1,     1,     1,      1, ...
     1,    2,     3,     4,     5,      6, ...
     4,   10,    18,    28,    40,     54, ...
    25,   74,   153,   268,   425,    630, ...
   208,  706,  1638,  3172,  5500,   8838, ...
  2146, 8162, 20898, 44164, 82850, 143046, ...
  ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened
A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.
Wikipedia, Feynman diagram
Wikipedia, Lattice path

Columns k=0-2 give: A005411 (for n>0), A000698(n+1), A005412(n+1).
Rows n=0-2 give: A000012, A000027(k+1), A028552(k+1).
Main diagonal gives A292693.
Cf. A258220, A258222.

b:= proc(x, y, t, k) option remember; `if`(y>x or y<0, 0,
      `if`(x=0, 1, b(x-1, y-1, false, k)*`if`(t, (x+k*y)/y, 1)
                 + b(x-1, y+1, true, k)  ))
    end:
A:= (n,k)-> b(2*n, 0, false, k):
seq(seq(A(n, d-n), n=0..d), d=0..12);

b[x_, y_, t_, k_] := b[x, y, t, k] = If[y>x || y<0, 0, If[x==0, 1, b[x-1, y -1, False, k]*If[t, (x+k*y)/y, 1] + b[x-1, y+1, True, k]]]; A[n_, k_] := b[2*n, 0, False, k]; Table[A[n, d-n], {d, 0, 12}, {n, 0, d}] // Flatten (* Jean-François Alcover, Jan 09 2016, after Alois P. Heinz *)

A(n,k) = Sum_{i=0..min(n,k)} C(k,i) * i! * A258220(n,i).

cp's OEIS Frontend

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A064801 Take 1, skip 2, take 2, skip 3, take 3, etc.

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