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Revision History for A324051 (Bold, blue-underlined text is an addition; faded, red-underlined text is a deletion.)

Showing entries 1-10 | older changes
#29 by Georg Fischer at Fri Mar 01 04:26:44 EST 2019
STATUS

proposed

approved

#28 by Antti Karttunen at Wed Feb 27 14:09:40 EST 2019
STATUS

editing

proposed

#27 by Antti Karttunen at Fri Feb 22 01:59:33 EST 2019
Discussion
Wed Feb 27
14:09
Antti Karttunen: (Please forget all the PinkBox-comments above. OK now for me).
#26 by Antti Karttunen at Thu Feb 21 15:32:05 EST 2019
COMMENTS

Any n which is of the form n = A005940(1+(2*(k^2))) for some k >= 1 [that is, a number that A156552 maps to a twice a square, thus n is odd by necessity], certainly cannot satisfy a(n) = 0, because no twice a square x can satisfy 0 = A106315(x) = x*d(x) mod sigma(x), as then x*d(x) will be even, and sigma(x) will be odd. E.g., 21 = A005940(1+(2*9)) certainly cannot be a solution to a(n) = 0. Numbers belonging that specific excluded set all satisfy A067029(n) = 1. But note that some of the solutions also satisfy it, e.g. 4199 = 13*17*19. - Antti Karttunen, Feb 21 2019

Discussion
Thu Feb 21
15:32
Antti Karttunen: Hah, too tired, got my odds and evens twisted.
15:35
Antti Karttunen: And in any case, such reasoning is vain, because the map A005940/A156552 is too well-lubricated, we just map some set of criteria to another set of criteria, but there's nothing (?) to tie between them, which would tighten constraints at either side of the map.
15:38
Antti Karttunen: Unless we can actually reduce such complex computation as A000203(A156552(n)) something not needing to compute A156552 first.
15:49
Antti Karttunen: In any case, such contrived reasoning "by periscope" is probably much easier in A324057 than here, as here we are swimming "against the grain". In A324057 numdiv is even converted to A106737 related to binomial identities.
#25 by Antti Karttunen at Thu Feb 21 15:29:10 EST 2019
COMMENTS

Any n which is of the form n = A005940(1+(2*(k^2))) for some k >= 1 [that is, a number that A156552 maps to a twice a square, thus n is odd by necessity], certainly cannot satisfy a(n) = 0, because no twice a square x can satisfy 0 = A106315(x) = 0, x*d(x) mod sigma(x), as then the expression x*d(x) at the left side of A106315 will be even, and sigma(x) at the right hand side will be odd. E.g., 21 = A005940(1+(2*9)) certainly cannot be a solution for to a(n) = 0 for that reason. Numbers belonging that specific excluded set all satisfy A067029(n) = 1. But note that some of the solutions also satisfy it, e.g. 4199 = 13*17*19 . - Antti Karttunen, Feb 21 2019

#24 by Antti Karttunen at Thu Feb 21 15:25:48 EST 2019
COMMENTS

Any n which is of the form n = A005940(1+(2*(k^2))) for some k >= 1 [that is, a number that A156552 maps to a twice a square, thus n is odd by necessity], certainly cannot satisfy a(n) = 0, because no twice a square x can satisfy A106315(x) = 0, as then the expression x*d(x) at the left side of A106315 will be even, and sigma(x) at the right hand side will be odd, (which as an obvious fact must be well known). E.g., 21 = A005940(1+(2*9)) cannot be a solution for a(n) = 0 for that reason. Numbers belonging that specific excluded set all satisfy A067029(n) = 1. But note that some of the solutions also satisfy it, e.g. 4199 = 13*17*19 - Antti Karttunen, Feb 21 2019

#23 by Antti Karttunen at Thu Feb 21 15:14:19 EST 2019
COMMENTS

Any n which is of the form n = A005940(1+(2*(k^2))) for some k >= 1 [that is, a number that A156552 maps to a twice a square, thus n is odd by necessity], certainly cannot satisfy a(n) = 0, because then the expression x*d(x) at the left side of A106315 will be even, and sigma(x) at the right hand side will be odd, (which as an obvious fact must be well known). E.g., 21 = A005940(1+(2*9)) cannot be a solution for a(n) = 0 for that reason. Numbers belonging that specific excluded set all satisfy A067029(n) = 1. But note that some of the solutions also satisfy it, e.g. 4199 = 13*17*19 - Antti Karttunen, Feb 21 2019

#22 by Antti Karttunen at Thu Feb 21 15:06:12 EST 2019
COMMENTS

Any n which is of the form n = A005940(1+(2*(k^2))) for some k >= 1 [that is, a number that A156552 maps to a twice a square, thus n is odd by necessity], certainly cannot satisfy a(n) = 0, because then the expression x*d(x) at the left side of A106315 will be even, and sigma(x) at the right hand side will be odd. E.g., 21 = A005940(1+(2*9)) cannot be a solution for a(n) = 0 for that reason. Numbers belonging that specific excluded set all satisfy A067029(n) = 1. But note that also some of the solutions also satisfy it, e.g. 4199 = 13*17*19 - Antti Karttunen, Feb 21 2019

#21 by Antti Karttunen at Thu Feb 21 15:05:05 EST 2019
COMMENTS

Any n which is of the form n = A005940(1+(2*(k^2))) for some k >= 1 [that is, a number that A156552 maps to a twice a square, thus n is odd by necessity], certainly cannot satisfy a(n) = 0, because then the expression x*d(x) at the left side of A106315 will be even, and sigma(x) at the right hand side will be odd. E.g., 21 = A005940(1+(2*9)) cannot be a solution for a(n) = 0 for that reason. Numbers belonging that specific excluded set all have satisfy A067029(n) = 1. But note that also some of the solutions also satisfy it, e.g. 4199 = 13*17*19 - Antti Karttunen, Feb 21 2019

#20 by Antti Karttunen at Thu Feb 21 15:02:46 EST 2019
COMMENTS

Any n which is of the form n = A005940(1+(2*(k^2))) for some k >= 1 [that is, a number that A156552 maps to a twice a square, thus n is odd by necessity], certainly cannot satisfy a(n) = 0, because then the expression x*d(x) at the left side of A106315 will be even, and sigma(x) at the right hand side will be odd. E.g., 21 = A005940(1+(2*9)) cannot be a solution for a(n) = 0 for that reason. Numbers belonging that specific excluded set all have A067029(n) = 1. - Antti Karttunen, Feb 21 2019