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COMMENTS
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The (2k+1)-sections A061037((2*k+1)*n+2) are multiples of 2k+1:
0,...21,...15,..117,...12,..285,...99,..525,...42,..837,..255, k=1, A142590
0,...45,...35,..285,...30,..725,..255,.1365,..110,.2205,..675, k=2, A165248
0,...77,...63,..525,...56,.1365,..483,.2597,..210,.4221,.1295, k=3, here
0,..117,...99,..837,...90,.2205,..783,.4221,..342,.6885,.2115, k=4,
0,..165,..143,.1221,..132,.3245,.1155,.6237,..506,10197,.3135, k=5
0,..221,..195,.1677,..182,.4485,.1599,.8645,..702,14157,.4355, k=6
After division by 2k+1 these define a table T'(k,c) :
0,....7,....5,...39,....4,...95,...33,..175,...14,..279,...85, k=1, A142883
0,....9,....7,...57,....6,..145,...51,..273,...22,..441,..135, k=2
0,...11,....9,...75,....8,..195,...69,..371,...30,..603,..185, k=3
0,...13,...11,...93,...10,..245,...87,..469,...38,..765,..235, k=4
0,...15,...13,..111,...12,..295,..105,..567,...46,..927,..285, k=5
0,...17,...15,..129,...14,..345,..123,..665,...54,.1089,..335, k=6
Differences downwards each second column in this second table are 2 = 7-5 = 9-7..; 18 = 57-39 = 75-57..; 50 = 145-95 = 195-145... = A077591(n+1) = 2*A016754.
The difference T'(k+1,c)-T'(k,c) is 0, 2, 2, 18, 2, 50, 18, 98, 8 ... = 2*A181318(c) =A061037(c-2)+A061037(c+2). - Paul Curtz, Mar 12 2012
Let b(n)= a(n) mod 11. The sequence b(n) has the property b(n+44) = b(n) with the first 43 values being {0, 0 , 8, 1, 1, 10, 1, 1, 8, 8, 0, 0, 10, 5, 5, 9, 7, 1, 5, 6, 10, 7, 0, 2, 1, 1, 8, 1, 5, 8, 2, 0, 8, 10, 6, 5, 10, 7, 9, 5, 0, 10, 0}. - G. C. Greubel, Apr 18 2016
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