A177083 - OEIS

%I #10 Mar 30 2012 18:52:05

%S 4,9,9,25,25,25,25,49,49,49,49,49,49,121,121,121,121,121,121,121,121,

%T 121,121,169,169,169,169,169,169,169,169,169,169,169,169

%N A006093(k)-fold repetition of A001248(k), k=1,2,3,..

%C Consider the initial terms of numerator sequences (dropping initial zeros) of

%C 3; A005563=N(1) ,

%C 5,3; A061037=N(2) ,

%C 7,16,1; A061039=N(3) ,

%C 9,5,33,3; A061041=N(4) ,

%C 11,24,39,56,3; A061043=N(5) ,

%C 13,7,5,4,85,1; A061045=N(6) ,

%C 15,32,51,72,95,120,3; A061047=N(7) ,

%C 17,9,57,5,105,33,161,3; A061049=N(8) ,

%C 19,40,7,88,115,16,175,208,1; N(9),

%C 21,11,69,6,1,39,189,14,261,3; N(10),

%C 23,48,75,104,135,168,203,240,279,320,3; N(11)

%C One must add the following associated (minimum) squares (taken from squared entries in A172038) to these values to reach the next possible square not larger than the entry itself:

%C 1;    N(1)

%C 4,1;    N(2)

%C 9,9,0;     N(3)

%C 16,4,16,1;    N(4)

%C 25,25,25,25,1;   N(5)

%C 36,9,4,0,36,0;    N(6)

%C 49,49,49,49,49,49,1;   N(7)

%C 64,16,64,4,64,16,64,1, ; N(8)

%C Only if the index of N(.) is a prime we obtain a string of equal consecutive terms in these complementary rows: 4, 9, 25, 49, 121, 169..

%C The current sequence lists the consecutive complementary squares, A001248, in the rows with prime index, including their multiplicity (which is A006093).

%C This generates a link between the primes and the Rydberg-Ritz spectrum of the hydrogen atom.

%Y Cf. A072055, A135177.

%K nonn,easy

%O 1,1

%A _Paul Curtz_, Dec 09 2010