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A125118 Triangle read by rows: T(n,k) = value of the n-th repunit in base (k+1) representation, 1<=k<=n. 11

%I

%S 1,3,4,7,13,21,15,40,85,156,31,121,341,781,1555,63,364,1365,3906,9331,

%T 19608,127,1093,5461,19531,55987,137257,299593,255,3280,21845,97656,

%U 335923,960800,2396745,5380840,511,9841,87381,488281,2015539,6725601

%N Triangle read by rows: T(n,k) = value of the n-th repunit in base (k+1) representation, 1<=k<=n.

%C T(n+1,k) = (k+1)*T(n,k) + 1;

%C row sums give A125120; central terms give A125119;

%C T(n,1) = A000225(n);

%C T(n,2) = A003462(n) for n>1;

%C T(n,3) = A002450(n) for n>2;

%C T(n,4) = A003463(n) for n>3;

%C T(n,5) = A003464(n) for n>4;

%C T(n,9) = A002275(n) for n>8;

%C T(n,n-2) = A031973(n) for n>2;

%C T(n,n-1) = A023037(n) for n>1;

%C T(n,n) = A060072(n+1);

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Repunit.html">Repunit</a>

%F T(n,k) = Sum((k+1)^i: 0<=i<n).

%e First 4 rows:

%e 1: [1]_2

%e 2: [11]_2 ........ [11]_3

%e 3: [111]_2 ....... [111]_3 ....... [111]_4

%e 4: [1111]_2 ...... [1111]_3 ...... [1111]_4 ...... [1111]_5

%e _

%e 1: 1

%e 2: 2+1 ........... 3+1

%e 3: (2+1)*2+1 ..... (3+1)*3+1 ..... (4+1)*4+1

%e 4: ((2+1)*2+1)*2+1 ((3+1)*3+1)*3+1 ((4+1)*4+1)*4+1

%e ((5+1)*5+1)*5+1.

%K nonn,tabl,base

%O 1,2

%A _Reinhard Zumkeller_, Nov 21 2006

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Last modified July 13 15:18 EDT 2020. Contains 335688 sequences. (Running on oeis4.)