A003015 -id:A003015 - cp's OEIS frontend

A100967 Least k such that binomial(2k + 1, k - n) >= binomial(2k, k).

3, 9, 18, 29, 44, 61, 81, 104, 130, 159, 191, 225, 263, 303, 347, 393, 442, 494, 549, 606, 667, 730, 797, 866, 938, 1013, 1091, 1172, 1255, 1342, 1431, 1524, 1619, 1717, 1818, 1922, 2029, 2138, 2251, 2366, 2485, 2606, 2730, 2857, 2987, 3119, 3255, 3394, 3535

Offset: 1

T. D. Noe, Nov 23 2004

nonn

From the formula, if we know k, we can estimate n as approximately 0.83 sqrt(k).

Open question: Does binomial(2*a(n) + 1, a(n) - n) = binomial(2*a(n), a(n)) for any n? An affirmative answer would settle whether there exists an odd term greater than 3 in A003016. - Danny Rorabaugh, Mar 16 2016

Charles R Greathouse IV, Table of n, a(n) for n = 1..500

Cf. A000984, A003015 (numbers that occur 5 or more times in Pascal's triangle).

F:= proc(n) local Q, LQ, k, k0;
     LQ:= -ln(GAMMA(k-n+1))-ln(GAMMA(k+1+n))-ln(k+1+n)+ln(2*k+1)+2*ln(GAMMA(k+1));
      k0:= floor(fsolve(LQ, k=n..max(2*n^2, 9)));
      if (2*k0+1)*binomial(k0, n) >= (n+1)*binomial(k0+1+n, n+1)  then
        while (2*k0-1)*binomial(k0-1, n) >= (n+1)*binomial(k0+n, n+1) do k0:= k0-1 od
      else
        while (2*k0+1)*binomial(k0, n) < (n+1)*binomial(k0+1+n, n+1) do k0:= k0+1 od;
      fi;
      k0;
end proc:
map(F, [$1..100]); # Robert Israel, Mar 16 2016

k=1; Table[While[Binomial[2k+1, k-n] < Binomial[2k, k], k++ ]; k, {n, 50}]

a(n,k=n+1)=while((2*k+1)*k!^2<(n+k+2)!*(k-n-1)!,k++);k \\ Charles R Greathouse IV, Sep 09 2013

Round(0.3807 + 1.43869 n + 1.44276 n^2) is an exact fit for the first 50 terms.

As n -> infinity, we have a(n) = (n^2+n)/log(2) + o(n). - Robert Israel, Mar 16 2016

A319382 Binomial coefficients binomial(m,k) for 2 <= k <= m/2 in sorted order.

Original entry on oeis.org

6, 10, 15, 20, 21, 28, 35, 36, 45, 55, 56, 66, 70, 78, 84, 91, 105, 120, 120, 126, 136, 153, 165, 171, 190, 210, 210, 220, 231, 252, 253, 276, 286, 300, 325, 330, 351, 364, 378, 406, 435, 455, 462, 465, 495, 496, 528, 560, 561, 595, 630, 666, 680, 703, 715, 741, 780, 792, 816, 820, 861, 903, 924

Offset: 1

Robert Israel, Sep 18 2018

nonn

In contrast to A006987, here the duplicates are not removed. Thus 120 = binomial(10,3) = binomial(16,2) appears twice.

			The first three terms are binomial(4,2) = 6, binomial(5,2) = 10, binomial(6,2) = 15.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A003015, A006987, A022911 (values of m), A022912 (values of k).

N:= 10^3: # to get terms <= N
Res:= NULL:
for n from 2 while n*(n-1)/2 <= N do
  for k from 2 to n/2 do
    v:= binomial(n,k);
    if v > N then break fi;
    Res:= Res,v
od od:
sort([Res]);

M = 10^3;
Reap[For[n = 2, n(n-1)/2 <= M, n++, For[k = 2, k <= n/2, k++, v = Binomial[n, k]; If[v > N, Break[]]; Sow[v]]]][[2, 1]] // Sort (* Jean-François Alcover, Apr 27 2019, from Maple *)

a(n) = binomial(A022911(n),A022912(n)).

A375573 Numbers occurring at least three times in Bernoulli's triangle A008949.

Original entry on oeis.org

1, 16, 64, 232, 256, 466, 562, 1024, 1486, 2048, 4096, 15226, 16384, 44552, 65536, 262144, 1048576, 4194304, 16777216, 67108864, 268435456, 1073741824, 4294967296, 17179869184, 68719476736, 274877906944, 1099511627776, 4398046511104, 17592186044416, 70368744177664

Offset: 1

Pontus von Brömssen, Aug 19 2024

nonn

Equivalently, 1 together with numbers occurring at least three times in columns k >= 1 of Bernoulli's triangle.
Equivalently, 1 together with numbers occurring at least twice in columns k >= 2 of Bernoulli's triangle.
4^j is a term if j >= 0 and j != 1, because 4^j = A008949(2*j,2*j) = A008949(2*j+1,j) = A008949(4^j-1,1) for j >= 2 and A008949(i,0) = 1 for all i. Are 232, 466, 562, 1486, 2048, 15226, 44552 the only terms not of this form? There are no more such terms below 2^70.
Are 1 and 4096 = A008949(12,12) = A008949(13,6) = A008949(90,2) = A008949(4095,1) the only numbers that occur at least 4 times? There are no more such numbers below 2^70.

Cf. A003015, A008949, A374796, A375570, A375572.

A062527 Smallest number (>1) which appears at least n times in Pascal's triangle.

Original entry on oeis.org

2, 3, 6, 10, 120, 120, 3003, 3003

Offset: 1

Henry Bottomley, Jul 10 2001

Singmaster's conjecture is that this sequence is finite.

			a(8)=3003 since 3003 =C(3003,1) =C(3003,3002) =C(78,2) =C(78,76) =C(15,5) =C(15,10) =C(14,6) = C(14,8).

H. L. Abbott, P. Erdos and D. Hanson, On the numbers of times an integer occurs as a binomial coefficient, Amer. Math. Monthly, (1974), 256-261.
FTL Magazine, One Thousand and One Coincidences
D. Singmaster, How often does an integer occur as a binomial coefficient?, Amer. Math. Monthly, 78 (1971), 385-386.
David Singmaster, Repeated binomial coefficients and Fibonacci numbers, Fibonacci Quarterly 13 (1975) 295-298.

Cf. A003015, A003016, A006987, A007318, A059233.

(* Computation lasts a few minutes *) max = 4000; Clear[cnt]; cnt[] = 0; Do[b = Binomial[n, k]; If[b <= max, cnt[b] += 1], {n, 0, max}, {k, 1, n - 1}]; sel = Select[Table[{b, cnt[b]}, {b, 1, max }], #[[2]] >= 1&]; a[n] := Select[sel, #[[2]] >= n&][[1, 1]]; Array[a, 8] (* Jean-François Alcover, Oct 05 2015 *)

A376001 Numbers that can be written as a Narayana number (A001263) in at least 3 ways.

Original entry on oeis.org

1, 105, 1176, 4950, 5713890

Offset: 1

Pontus von Brömssen, Sep 06 2024

The first 5 terms are triangular numbers.
a(2), ..., a(5) can all be written as a Narayana number in exactly 4 ways.
a(6) > 2*10^35 (if it exists).

			With T(n,k) = A001263(n,k):
      105 = T( 7,3) = T( 7, 5) = T(  15,2) = T(  15,  14);
     1176 = T( 9,4) = T( 9, 6) = T(  49,2) = T(  49,  48);
     4950 = T(11,4) = T(11, 8) = T( 100,2) = T( 100,  99);
  5713890 = T(92,3) = T(92,90) = T(3381,2) = T(3381,3380).

Cf. A000217, A001263, A003015, A374796, A375573, A375999, A376000.

from math import isqrt
from bisect import insort
from itertools import islice
def A010054(n):
    return isqrt(m:=8*n+1)**2 == m
def A376001_generator():
    yield 1
    nkN_list = [(5, 3, 20)] # List of triples (n, k, A001263(n, k)), sorted by the last element.
    while 1:
        N0 = nkN_list[0][2]
        c = 0
        while 1:
            n, k, N = nkN_list[0]
            if N > N0:
                if c >= 3 or A010054(N0): yield N0
                break
            central = n==2*k-1
            c += 2-central
            del nkN_list[0]
            insort(nkN_list, (n+1, k, n*(n+1)*N//((n-k+1)*(n-k+2))), key=lambda x:x[2])
            if central:
                insort(nkN_list, (n+2, k+1, 4*n*(n+2)*N//(k+1)**2), key=lambda x:x[2])
def A376001_list(nmax):
    return list(islice(A376001_generator(),nmax))

A138496 Where record values occur in A003016.

Original entry on oeis.org

0, 1, 10, 120, 3003

Offset: 1

Reinhard Zumkeller, Mar 20 2008

It appears that the record values are 0, 3, 4, 6, 8, 10, 12, ...
From M. F. Hasler, Feb 16 2023: (Start)
The numbers that appear 3 times in Pascal's triangle are the central binomial coefficients (A000984), except for the number 2 that is the only number to appear only once. For n = 1 there would be an infinite number of occurrences, but sequence A003016 counts only the occurrences of n in rows <= n so that n = 1 also gives 3.
All C(n,k) with 1 < k < n/2 (in particular triangular numbers A000217) appear at least 4 times; see A098564 for those appearing exactly 4 times.
Numbers that appear 5 or more times are quite rare, they are listed in A003015 with subsequence A098565 of those appearing exactly 6 times.
They are mostly C(n,k) with 2 < k < n/2 which are also triangular numbers, but some are also of the form C(n+1,k) = C(n,k+1) with 3 < k < n/2, and a subsequence of these has n and k given in terms of Fibonacci numbers. (End)

m=-1; [n | n<-[0..9999], m < m = max(A003016(n), m)] \\ M. F. Hasler, Feb 16 2023

m=-1; [n for n in range(9999)if m < (m := max(A003016(n), m))] # M. F. Hasler, Feb 16 2023

A270440 Least k such that binomial(k, 2) >= binomial(2*n, n).

Original entry on oeis.org

2, 3, 4, 7, 13, 23, 44, 84, 161, 313, 609, 1189, 2327, 4562, 8958, 17614, 34673, 68318, 134724, 265878, 525066, 1037554, 2051390, 4057939, 8030892, 15900354, 31493446, 62400953, 123682583, 245223436, 486342641, 964809156, 1914483817, 3799849586, 7543612064, 14979070587, 29749371096, 59095356237, 117410567231

Offset: 0

Danny Rorabaugh, Mar 17 2016

nonn

Open question: Does binomial(a(n), 2) = binomial(2*n, n) for any n > 2? An affirmative answer would settle whether there exists an odd term greater than 3 in A003016.

binomial(a(n),2) > binomial(2*n,n) for 2 < n <= 800000. - Chai Wah Wu, Mar 22 2016

Chai Wah Wu, Table of n, a(n) for n = 0..1000

Cf. A000217, A000984, A003015, A100967.

Table[SelectFirst[Range[10^7], Binomial[#, 2] >= Binomial[2 n, n] &], {n, 0, 22}] (* Michael De Vlieger, Mar 17 2016, Version 10 *)

a(n) = {my(c = binomial(2*n, n)); my(k = 0); while (binomial(k,2) < c, k++); k;} \\ Michel Marcus, Mar 17 2016

from _future_ import division
from gmpy2 import iroot
A270440_list, b = [], 8
for n in range(1001):
    q, r = iroot(b+1,2)
    A270440_list.append(int((q+1)//2 + (0 if r else 1)))
    b = b*2*(2*n+1)//(n+1) # Chai Wah Wu, Mar 22 2016

def k2_2nn(M): # Produces the first M terms.
    K, n, center, k, triangle = [], 0, 1, 1, 0
    while len(K)

Conjecture: a(n) ~ 2^(n + 1/2) / (Pi*n)^(1/4). - Vaclav Kotesovec, Mar 23 2016

a(n) = ceiling(((8*binomial(2*n,n)+1)^(1/2)+1)/2). The above conjecture is true asymptotically. Using Stirling's formula for the approximation of n!, we get binomial(2*n,n) ~ 2^(2*n)/(Pi*n)^(1/2) and inserting this in the formula for a(n) results in the above approximation for a(n). - Chai Wah Wu, Mar 23 2016

A128373 Irregular triangle read by rows: row n (n>=2) lists positions in the sequence A007318 where n appears.

Original entry on oeis.org

4, 7, 8, 11, 13, 16, 19, 12, 22, 26, 29, 34, 37, 43, 46, 53, 17, 18, 56, 64, 67, 76, 79, 89, 92, 103, 106, 118, 23, 25, 121, 134, 137, 151, 154, 169, 172, 188, 191, 208, 24, 211, 229, 30, 33, 232, 251, 254, 274, 277, 298, 301, 323, 326, 349, 352, 376, 379, 404, 38

Offset: 2

Nick Hobson, Mar 01 2007

			In A007318, the number 2 appears in position 4, so the first row is 4. The number 3 appears in positions 7 and 8 in A007318, so the second row is 7, 8.
The irregular triangle begins:
   4
   7,  8
  11, 13
  16, 19
  12, 22, 26
  29, 34
  37, 43
  46, 53
  ...

Cf. A003015, A003016, A007318, A059233.

A298472 Numbers n such that n and n-1 are both nontrivial binomial coefficients.

Original entry on oeis.org

21, 36, 56, 253, 496, 561, 1771, 2926, 3655, 5985, 26335, 2895621, 2919736, 6471003, 21474181, 48792381, 346700278, 402073903, 1260501229261, 12864662659597529

Offset: 1

Christopher E. Thompson, Jan 19 2018

nonn

Nontrivial here means binomial(r,s) with 2 <= s <= r-2 (or the sequence would be uninteresting).

Blokhuis et al. show that the values given are complete up to 10^30, and conjecture that there are no more.

			binomial(6,3)=20 and binomial(7,2)=binomial(7,5)=21 are the smallest adjacent pair, so a(1)=21.

Aart Blokhuis, Andries Brouwer, Benne de Weger, Binomial collisions and near collisions, INTEGERS, Volume 17, Article A64, 2017 (also available as arXiv:1707.06893 [math.NT]).

Cf. A003015.

nmax = 1000; t = Table[Binomial[n, k], {n, 4, nmax}, {k, 2, Floor[n/2]}] // Flatten // Sort // DeleteDuplicates; Select[Split[t, #2 == #1+1&], Length[#] > 1&][[All, 2]] (* Jean-François Alcover, Feb 20 2018 *)

cp's OEIS Frontend

A100967 Least k such that binomial(2k + 1, k - n) >= binomial(2k, k).

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A319382 Binomial coefficients binomial(m,k) for 2 <= k <= m/2 in sorted order.

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A375573 Numbers occurring at least three times in Bernoulli's triangle A008949.

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A062527 Smallest number (>1) which appears at least n times in Pascal's triangle.

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A376001 Numbers that can be written as a Narayana number (A001263) in at least 3 ways.

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A138496 Where record values occur in A003016.

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PARI

Python

A270440 Least k such that binomial(k, 2) >= binomial(2*n, n).

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A128373 Irregular triangle read by rows: row n (n>=2) lists positions in the sequence A007318 where n appears.

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A298472 Numbers n such that n and n-1 are both nontrivial binomial coefficients.

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