0,4

A variant of the triple factorial array A136212. Compare to triangle array A136218, which is generated by a complementary process.

Let triangular matrix P = A136220, then: column 0 (A136221) = column 0 of P; column 1 (A136226) = column 0 of P^2; column 3 (A136229) = column 0 of P^4.

Square array begins:

(1),(1),1,(1),1,(1),1,(1),1,1,(1),1,1,(1),1,1,(1),1,1,(1),1,1,1,(1),...;

(1),(2),3,(4),5,(6),7,(8),9,10,(11),12,13,(14),15,16,(17),18,19,20,(21),..;

(3),(8),15,(24),34,(46),59,(74),90,108,(127),147,169,(192),216,242,(269),..;

(15),(49),108,(198),306,(453),622,(838),1080,1377,(1704),2062,2485,(2943),..;

(108),(414),1036,(2116),3493,(5555),8040,(11477),15483,20748,(26748),33528,..;

(1036),(4529),12569,(28052),48800,(82328),124335,(186261),260856,364551,..;

(12569),(61369),185704,(446560),811111,(1438447),2250731,(3513569),5078154,..;

(185704),(996815),3247546,(8325700),15684001,(29039188),46830722,...;

(3247546),(18931547),65762269,(178284892),346583419,...;

(65762269),(412345688),1515642725,(4317391240),...; ...

where terms in parenthesis are at positions {[m*(m+7)/6], m>=0}

and are removed before taking partial sums to obtain the next row.

To generate the array, start with all 1's in row 0; from then on,

obtain row n+1 from row n by first removing terms in row n at

positions {[m*(m+7)/6], m>=0} and then taking partial sums.

For example, to generate row 2 from row 1:

[(1),(2),3,(4),5,(6),7,(8),9,10,(11),12,13,(14),15,16,(17),18,...],

remove terms at positions [0,1,3,5,7,10,13,16,20,...] to get:

[3, 5, 7, 9,10, 12,13, 15,16, 18,19,20, 22,23,24, 26,27,28,...]

then take partial sums to obtain row 2:

[3,8,15,24,34,46,59,74,90,108,127,147,169,192,216,242,269,...].

Continuing in this way will generate all the rows of this array.

Amazingly, column 0 of this array = column 0 of triangle P=A136220:

1;

1, 1;

3, 2, 1;

15, 10, 3, 1;

108, 75, 21, 4, 1;

1036, 753, 208, 36, 5, 1;

12569, 9534, 2637, 442, 55, 6, 1;

185704, 146353, 40731, 6742, 805, 78, 7, 1; ...

where column k of P^3 = column 0 of P^(3k+3) such that

column 0 of P^3 = column 0 of P shift one place left.

(PARI) {T(n, k)=local(A=0, m=0, c=0, d=0); if(n==0, A=1, until(d>k, if(c==(m*(m+7))\6, m+=1, A+=T(n-1, c); d+=1); c+=1)); A}

Cf. columns: A136221, A136226, A136229; related tables: A136220 (P), A136226 (P^2), A136232 (P^4).

Adjacent sequences: A136214 A136215 A136216 this_sequence A136218 A136219 A136220

Sequence in context: A109282 A135902 A135876 this_sequence A136220 A088956 A106208

nice,nonn,tabl

Paul D. Hanna (pauldhanna(AT)juno.com), Dec 23 2007