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(100^n,1) Pascal triangle
4

%I #9 Nov 02 2016 15:59:54

%S 1,100,1,10000,101,1,1000000,10101,102,1,100000000,1010101,10203,103,

%T 1,10000000000,101010101,1020304,10306,104,1,1000000000000,

%U 10101010101,102030405,1030610,10410,105,1,100000000000000,1010101010101

%N (100^n,1) Pascal triangle

%C (For row n=0,1,2,3,4,5 : Sum of digits = Pascal triangle)

%F T(n,0)=100^n, T(n,n)=1, T(n,k)=T(n-1,k-1)+T(n-1,k) for 0<k<n. - _Philippe Deléham_, Dec 27 2013

%F T(n,k)=101*T(n-1,k)+T(n-1,k-1)-100*T(n-2,k)-100*T(n-2,k-1), T(0,0)=1, T(1,0)=100, T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - _Philippe Deléham_, Dec 27 2013

%e Triangle begins:

%e 1

%e 100,1

%e 10000,101,1

%e 1000000,10101,102,1

%e 100000000,1010101,10203,103,1

%e 10000000000,101010101,1020304,10306,104,1

%e 1000000000000,10101010101,102030405,1030610,10410,105,1

%e 100000000000000,1010101010101,10203040506,103061015,1041020,10515,106,1

%e 10000000000000000,101010101010101,1020304050607,10306101521,104102035,1051535,10621,107,1,

%Y Cf. A228196

%K nonn,tabl,easy

%O 0,2

%A _Mark Dols_, Aug 28 2009