A003016 -id:A003016 - cp's OEIS frontend

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

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 *)

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

A094011 a(n) = number of occurrences of n as a trinomial coefficient (in A046816).

Original entry on oeis.org

3, 6, 6, 6, 10, 6, 6, 6, 12, 6, 9, 6, 6, 12, 6, 6, 6, 6, 12, 12, 6, 6, 6, 6, 6, 6, 12, 6, 12, 6, 6, 6, 6, 12, 12, 6, 6, 6, 6, 6, 9, 6, 6, 12, 6, 6, 6, 6, 6, 6, 6, 6, 6, 12, 15, 6, 6, 6, 12, 6, 6, 6, 6, 6, 12, 6, 6, 6, 9, 6, 9, 6, 6, 6, 6, 6, 12, 6, 6, 6, 6, 6, 12, 6, 6, 6, 6, 6, 10, 12, 6, 6, 6, 6, 6, 6, 6

Offset: 2

Lior Manor, Apr 21 2004

nonn

a(n) <= 3 * A003016(n).

			a(2) = 3 since, 2 appears only 3 times in A046816.

Cf. A046816, A003016.

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.

A158086 Number of occurrences of n as an entry in rows <= 2n of Losanitsch's triangle (A034851).

Original entry on oeis.org

4, 4, 5, 4, 6, 4, 4, 6, 5, 4, 6, 4, 4, 4, 6, 4, 4, 6, 6, 4, 4, 4, 4, 6, 4, 4, 6, 4, 6, 4, 4, 4, 4, 4, 6, 4, 5, 4, 4, 4, 6, 4, 6, 4, 4, 4, 4, 6, 4

Offset: 2

Alonso del Arte, Mar 12 2009

For n = 1 to 1000, the only values of a(n) are 4, 5, 6, 8, 10 and infinity.

			a(4) = 5 because 4 occurs five times in Losanitsch's triangle: the first time at row 4, column 2, being the sum of the two 2's in the row above; and at column 1 of rows 7 and 8, which are symmetrically duplicated at row 7, column 6 and row 8, column 7.

Cf. A003016, Number of occurrences of n as an entry in rows <= n of Pascal's triangle.

(* The following assumes a[n, k] has already been defined to give Losanitsch's triangle; see for example the program given for A153046 *)
tallyLozOccs[1] := Infinity; tallyLozOccs[n_Integer?Positive] := Module[{i, searchMax, tally}, i = 0; searchMax = 2n; tally = 0; While[i <= searchMax, tally = tally + Length[Select[Table[a[i, m], {m, 0, i}], # == n &]]; i++ ]; Return[tally]]; Table[tallyLozOccs[n], {n, 2, 50}]
(* this program also assumes a(n,k) has been defined for Losanitsch's triangle*)
Table[Length[Select[Flatten[Table[a[i,m], {i,0,2n}, {m,0,i}]],#==n&]], {n,2,50}] (* Wilfredo Lopez (chakotay147138274(AT)yahoo.com), Mar 18 2009 *)

A260466 Number of integers in Pascal's triangle strictly between 1 and n.

Original entry on oeis.org

0, 0, 1, 3, 5, 7, 10, 12, 14, 16, 20, 22, 24, 26, 28, 32, 34, 36, 38, 40, 43, 47, 49, 51, 53, 55, 57, 59, 63, 65, 67, 69, 71, 73, 75, 79, 83, 85, 87, 89, 91, 93, 95, 97, 99, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 125, 129, 131, 133, 135, 137, 139, 141

Offset: 1

Alex Jordan, Jul 26 2015

nonn

Ignoring the first two terms of A003016, a(n) is partial sums of A003016.

a(n) >= 2n-5 trivially; for n>=7, a(n) > 2n-5.

			For n=7, the members of Pascal's triangle strictly between 1 and 7 are C(2,1), C(3,1), C(3,2), C(4,1), C(4,2), C(4,3), C(5,1), C(5,4), C(6,1), and C(6,5). So a(7)=10.

Cf. A003016, A007318.

t = 0 * Range[101]; Do[x = Binomial[a, b]; If[1 < x <= 100, t[[x + 1]]++], {a, 100}, {b, a}]; Accumulate@ t (* Giovanni Resta, Aug 16 2015 *)

nbn(n) = {my(nb = 0); for (j=1, n, for (k=1, n, b = binomial(j, k); if ((b>1) && (b<=n), nb++););); nb;} \\ Michel Marcus, Jul 30 2015

More terms from Michel Marcus, Jul 30 2015

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

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

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A094011 a(n) = number of occurrences of n as a trinomial coefficient (in A046816).

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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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A158086 Number of occurrences of n as an entry in rows <= 2n of Losanitsch's triangle (A034851).

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Mathematica

A260466 Number of integers in Pascal's triangle strictly between 1 and n.

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