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A194705 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (5 + m). 2
7, 4, 3, 2, 2, 3, 1, 1, 3, 2, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 2, 0, 1, 0, 1, 1, 2, 2, 0, 0, 1, 0, 1, 1, 2, 2, 0, 0, 0, 1, 0, 1, 1, 2, 2, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2 (list; table; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

Sub-triangle of A182703 and also of A194812. Note that the sum of every row is also the number of partitions of 5. For further information see A182703 and A135010.

LINKS

Table of n, a(n) for n=1..91.

FORMULA

T(k,m) = A182703(5+m,k), with T(k,m) = 0 if k > 5+m.

T(k,m) = A194812(5+m,k).

EXAMPLE

Triangle begins:

7,

4, 3,

2, 2, 3,

1, 1, 3, 2,

0, 1, 1, 2, 3,

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

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

...

For k = 1 and  m = 1; T(1,1) = 7 because there are seven parts of size 1 in the last section of the set of partitions of 6, since 5 + m = 6, so a(1) = 7. For k = 2 and m = 1; T(2,1) = 4 because there are four parts of size 2 in the last section of the set of partitions of 6, since 5 + m = 6, so a(2) = 4.

PROG

(PARI) P(n)={my(M=matrix(n, n), d=5); M[1, 1]=numbpart(d); for(m=1, n, forpart(p=m+d, for(k=1, #p, my(t=p[k]); if(t<=n && m<=t, M[t, m]++)), [2, m+d])); M}

{ my(T=P(10)); for(n=1, #T, print(T[n, 1..n])) } \\ Andrew Howroyd, Feb 19 2020

CROSSREFS

Always the sum of row k = p(5) = A000041(5) = 7.

The first (0-10) members of this family of triangles are A023531, A129186, A194702-A194704, this sequence, A194706-A194710.

Cf. A135010, A138121, A194812.

Sequence in context: A198348 A019857 A296474 * A344906 A243309 A244817

Adjacent sequences:  A194702 A194703 A194704 * A194706 A194707 A194708

KEYWORD

nonn,tabl

AUTHOR

Omar E. Pol, Feb 05 2012

EXTENSIONS

Terms a(29) and beyond from Andrew Howroyd, Feb 19 2020

STATUS

approved

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Last modified December 5 00:21 EST 2021. Contains 349530 sequences. (Running on oeis4.)